If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on Stoichiometry

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Balance Chemical Reactions
(2.) Measurements and Units

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.

(1.) WASSCE In a neutralization reaction, dilute tetraoxosulphate $(VI)$ acid completely reacted with sodium hydroxide solution.
(i) Write a balanced equation for the reaction.
(ii) How many moles of sodium hydroxide would be required for the complete neutralization of $0.50$ moles of tetraoxosulphate $(VI)$ acid?


$ tetraoxosulphate\:(VI)\:acid = H_2SO_4 \\[3ex] sodium\:\:hydroxide = NaOH \\[3ex] H_2SO_4 + NaOH \rightarrow Na_2SO_4 + H_2O...neutralization\:\:reaction \\[3ex] Balance\:\:the\:\:equation \\[3ex] (i) \\[3ex] H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \\[3ex] (ii) \\[3ex] Based\:\:on\:\:the\:\:balanced\:\:equation \\[3ex] 2\:moles\:\:of\:\:NaOH\:\:neutralizes\:\:1\:\:mole\:\:H_2SO_4 \\[3ex] $ Let the number of moles of sodium hydroxide that would be required for the complete neutralization of $0.50$ moles of tetraoxosulphate $(VI)$ acid be $p$
Proportional Reasoning Method
$H_2SO_4$ (moles) $NaOH$ (moles)
$1$ $2$
$0.5$ $p$

$ Cross\:\:Multiply \\[3ex] p * 1 = 0.5 * 2 \\[3ex] p = 1\:mole \\[3ex] $ $1$ mole of sodium hydroxide would be required for the complete neutralization of $0.50$ moles of tetraoxosulphate $(VI)$ acid
(2.) Given the reaction: $2Al_2O_3 \rightarrow 4Al + 3O_2$
How many grams of $Al$ would be obtained from $10$ moles of $Al_2O_3$
$(Al = 27,\:\:O = 16)$


$ 2Al_2O_3 \rightarrow 4Al + 3O_2 \\[3ex] 2\:moles\:\:of\:\:Al_2O_3 \:\:gives\:\: 4\:moles\:\:of\:\:Al \\[3ex] 1\:mole \:\:of\:\:Al = 27g \\[3ex] \therefore 4\:moles \:\:of\:\:Al = 4 * 27 = 108g \\[3ex] \implies 2\:moles\:\:of\:\:Al_2O_3 \:\:gives\:\: 108\:grams\:\:of\:\:Al \\[3ex] $ Let the number of grams of $Al$ that would be obtained from $10$ moles of $Al_2O_3$ be $p$
Proportional Reasoning Method
$Al_2O_3$ (moles) $Al$ (grams)
$2$ $108$
$10$ $p$

$ Cross\:\:Multiply \\[3ex] p * 2 = 10 * 108 \\[3ex] p = \dfrac{10 * 108}{2} \\[5ex] p = 5 * 108 \\[3ex] p = 540\:grams \\[3ex] $ $540$ grams of $Al$ would be obtained from $10$ moles of $Al_2O_3$
(3.) A $35.0\:mL$ sample of $HCl$ of unknown concentration is titrated to its end point with $19.6\:mL$ of $0.50\:M$ $NaOH$.
What is the molarity of the $HCl$?


$ HCl + NaOH \rightarrow NaCl + H_2O...neutralization\:\:reaction...balanced \\[3ex] 1\:mol\:\:HCl \:\:reacts\:\:with\:\:1\:\:mol\:\:NaOH \\[3ex] \dfrac{C_A * V_A}{C_B * V_B} = \dfrac{n_A}{n_B} \\[5ex] C_A = ? \\[3ex] V_A = 35\:mL \\[3ex] C_B = 0.5\:M \\[3ex] V_B = 19.6\:mL \\[3ex] n_A = 1\:mol \\[3ex] n_B = 1\:mol \\[3ex] \dfrac{C_A * 35}{0.5 * 19.6} = \dfrac{1}{1} \\[5ex] Cross\:\:Multiply \\[3ex] 1(C_A * 35) = 1(0.5 * 19.6) \\[3ex] C_A * 35 = 0.5 * 19.6 \\[3ex] C_A = \dfrac{0.5 * 19.6}{35} \\[5ex] C_A = \dfrac{9.8}{35} \\[5ex] C_A = 0.28\:M \\[3ex] $ The molarity of the $HCl$ is $0.28\:M$
(4.) Calculate the molarity of a solution that contains $30.0$ grams of calcium chloride in $1.2$ liters of solution.
$(Ca = 40,\:\: Cl = 35.5)$


$ Molarity = \dfrac{concentration\:\:in\:\:moles}{volume\:\:in\:\:dm^3} \\[5ex] Molarity = \dfrac{mol}{dm^3} \\[5ex] 1\:dm^3 = 1\:L \\[3ex] Molarity = \dfrac{mol}{L} \\[5ex] 1\:mole\:\:of\:\:CaCl_2 = 40 + (35.5 * 2) \\[3ex] = 40 + 71 \\[3ex] = 111g \\[3ex] 1\:mole\:\:of\:\:CaCl_2 = 111g \\[3ex] how\:\:many\:\:moles = 30g? \\[3ex] Let\:\:the\:\:number\:\:of\:\:moles = x \\[3ex] $
Proportional Reasoning Method
$CaCl_2$ (moles) $CaCl_2$ (grams)
$1$ $111$
$x$ $30$

$ Cross\:\:Multiply \\[3ex] x * 111 = 1 * 30 \\[3ex] x * 111 = 30 \\[3ex] x = \dfrac{30}{111} \\[5ex] x = 0.2702702703\:moles \\[3ex] volume = 1.2L \\[3ex] \implies Molarity = \dfrac{0.2702702703}{1.2} \\[5ex] Molarity = 0.2252252252M $
(5.)


$ t = 4\:years \\[3ex] Semiannual\:\:bond \rightarrow m = 2 \\[3ex] FV = \$1000 \\[3ex] BCR = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] YTM = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.06}{0.07} \\[5ex] = \dfrac{60}{0.07} \\[5ex] = 857.142857 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 2(30) = 60 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.07}{2} = 1 + 0.035 = 1.035 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.035)^{60} = 7.8780909 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{7.8780909} = 0.126934306 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.126934306 = 0.873065694 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{7.8780909} \\[5ex] = 126.934306 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 857.142857 * 0.873065694 + 126.934306 \\[3ex] BP = 748.342023 + 126.934306 \\[3ex] BP = 875.276329 \\[3ex] BP \approx \$875.28 $
(6.) Given the reaction: $2Al_2O_3 \rightarrow 4Al + 3O_2$
How many grams of $Al$ would be obtained from $5.8$ grams of $Al_2O_3$
$(Al = 27,\:\:O = 16)$


$ 2Al_2O_3 \rightarrow 4Al + 3O_2 \\[3ex] 2\:moles\:\:of\:\:Al_2O_3 \:\:gives\:\: 4\:moles\:\:of\:\:Al \\[3ex] 1\:mole\:\:of\:\:Al_2O_3 = (27 * 2) + (16 * 3) = 54 + 48 = 102\:grams \\[3ex] \therefore 2\:moles \:\:of\:\:Al_2O_3 = 2 * 102 = 204g \\[3ex] 1\:mole \:\:of\:\:Al = 27g \\[3ex] \therefore 4\:moles \:\:of\:\:Al = 4 * 27 = 108g \\[3ex] \implies 204\:grams\:\:of\:\:Al_2O_3 \:\:gives\:\: 108\:grams\:\:of\:\:Al \\[3ex] $ Let the number of grams of $Al$ that would be obtained from $5.8$ grams of $Al_2O_3$ be $p$
Proportional Reasoning Method
$Al_2O_3$ (grams) $Al$ (grams)
$204$ $108$
$5.8$ $p$

$ Cross\:\:Multiply \\[3ex] p * 204 = 5.8 * 108 \\[3ex] p = \dfrac{5.8 * 108}{204} \\[5ex] p = \dfrac{626.4}{204} \\[5ex] p = 3.070588235\:grams \\[3ex] p \approx 3.1\:g \\[3ex] $ About $3.1$ grams of $Al$ would be obtained from $5.8$ grams of $Al_2O_3$
(7.) GCSE
$(7.1)$ In Stage 2, $40\;kg$ of titanium chloride was added to $20\;kg$ of sodium.
The equation for the reaction is:
$TiCl_4 + 4Na \rightarrow Ti + 4NaCl$
Relative atomic masses $(A_r):\;\; Na = 23\;\; Cl = 35.5\;\; Ti = 48$
Explain why titanium chloride is the limiting reactant.
You must show your working.
$(7.2)$ For a Stage 2 reaction, the percentage yield was $92.3\%$
The theoretical maximum mass of titanium produced in this batch was $12.5\;kg$
Calculate the actual mass of titanium produced.


$ (7.1) \\[3ex] TiCl_4 + 4Na \rightarrow Ti + 4NaCl \\[3ex] \underline{Supposed\;\;to} \\[3ex] 1\;mol\;TiCl_4 \rightarrow 4\;mol\;Na \\[3ex] 1\;mol\;TiCl_4 = 48 + (35.5 * 4) = 48 + 142 = 190\;g \\[3ex] 4\;mol\;Na = 4 * 23 = 92\;g \\[3ex] \implies 190\;g\;TiCl_4 \rightarrow 92\;g\;Na \\[3ex] \underline{Given} \\[3ex] 40\;kg\;TiCl_4 = 40 * 1000 = 40000\;g\;TiCl_4 \\[3ex] 20\;kg\;Na = 20 * 1000 = 20000\;g\;Na \\[3ex] $
Table 1 (Find amount of sodium)
$TiCl_4\;(g)$ $Na\;(g)$
$190$ $92$
$40000$ $x$

$ \dfrac{x}{92} = \dfrac{40000}{190} \\[5ex] x = \dfrac{92 * 40000}{190} \\[5ex] x = 19368.42105\;g\;Na ...required \\[3ex] But\;\;given:\;\; 20000\;g\;Na \\[3ex] given \gt required \implies Sodium\;\;is\;\;excess\;\;reactant \\[3ex] $
Table 2 (Find amount of titanium chloride)
$TiCl_4\;(g)$ $Na\;(g)$
$190$ $92$
$y$ $20000$

$ \dfrac{y}{20000} = \dfrac{190}{92} \\[5ex] y = \dfrac{20000 * 190}{92} \\[5ex] y = 41304.34783\;g\;TiCl_4 ...required \\[3ex] But\;\;given:\;\; 40000\;g\;TiCl_4 \\[3ex] given \lt required \implies Titanium\;chloride\;\;is\;\;limiting\;\;reactant \\[5ex] (7.2) \\[3ex] Percentage\;\;Yield = \dfrac{Actual\;\;Yield}{Theoretical\;\;Yield} * 100 \\[5ex] 92.3\% = Percentage\;\;Yield = \dfrac{Actual\;\;Yield}{13.5} \\[5ex] Swap \\[3ex] \dfrac{Actual\;\;Yield}{13.5} = 92.3\% \\[5ex] \dfrac{Actual\;\;Yield}{13.5} = \dfrac{92.3}{100} \\[5ex] Actual\;\;Yield = \dfrac{13.5 * 92.3}{100} \\[5ex] Actual\;\;Yield = 12.4605\;kg \approx 12.5\;kg $
(8.) WASSCE Calcium carbonate of mass $1.0\;g$ was heated until there was no further change.
(i) Write an equation for the reaction which took place.
(ii) Calculate the mass of the residue.
(iii) Calculate the volume of the gas at s.t.p
(iv) What would be the volume of the gas measured at $15^\circ C$ and $760\;mm\;Hg$?
$[C = 12.0,\;\; O = 16.0,\;\; Ca = 40.0]$
Molar volume of a gas at s.t.p = $22.4\;dm^3$


$ (i) \\[3ex] {CaCO_3}_{(s)} \rightarrow {CaO}_{(s)} + {CO_2}_{(g)} \\[3ex] (ii) \\[3ex] 1\;mol\;CaCO_3 \rightarrow 1\;mol\;CaO \\[3ex] 1\;mol\;CaCO_3 = 40 + 12 + (16 * 3) = 52 + 48 = 100\;g \\[3ex] 1\;mol\;CaO = 40 + 16 = 56\;g \\[3ex] \implies 100\;g\;CaCO_3 \rightarrow 56\;g\;CaO \\[3ex] $
Proportional Reasoning Method
$CaCO_3\;(grams)$ $CaO\;(grams)$
$100$ $56$
$1$ $m$

$ \dfrac{m}{1} = \dfrac{56}{100} \\[5ex] m = 0.56\;g \\[3ex] $ $0.56\;g$ of calcium oxide is produced from the decomposition of $1.0\;g$ of calcium trioxocarbonate $(IV)$

$ (iii) \\[3ex] 1\;mol\;CaCO_3 \rightarrow 1\;mol\;CO_2 \\[3ex] 1\;mol\;CO_2\;\;occupies\;\;22.4\;dm^3\;\;at\;\;s.t.p \\[3ex] \implies 1\;mol\;CaCO_3 \rightarrow 22.4\;dm^3\;CO_2\;\;at\;\;s.t.p \\[3ex] \implies 100\;g\;CaCO_3 \rightarrow 22.4\;dm^3\;CO_2\;\;at\;\;s.t.p \\[3ex] $
Proportional Reasoning Method
$CaCO_3\;(grams)$ $CO_2\;(dm^3)$
$100$ $22.4$
$1$ $v$

$ \dfrac{v}{1} = \dfrac{22.4}{100} \\[5ex] v = 0.224\;dm^3 \\[3ex] $ The decomposition of $1.0\;g$ of calcium trioxocarbonate $(IV)$ leads to the emission of $0.224\;dm^3$ of carbon $(IV)$ oxide at standard temperature and pressure.

$ (iv) \\[3ex] P_1 = 760\;mm\;Hg \\[3ex] V_1 = 0.224\;dm^3 \\[3ex] T_1 = 273\;K \\[3ex] P_2 = 760\;mm\;Hg \\[3ex] V_2 = ? \\[3ex] T_2 = 15^\circ C = 15 + 273 = 288\;K \\[3ex] \underline{General\;\;Gas\;\;Equation} \\[3ex] \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] \dfrac{760 * 0.224}{273} = \dfrac{760 * V_2}{288} \\[5ex] \dfrac{0.224}{273} = \dfrac{V_2}{288} \\[5ex] Swap \\[3ex] \dfrac{V_2}{288} = \dfrac{0.224}{273} \\[5ex] V_2 = \dfrac{288 * 0.224}{273} \\[5ex] V_2 = 0.2363076923\;dm^3 \\[3ex] $ The volume of the gas $CO_2$ measured at $15^\circ C$ and $760\;mm\;Hg$ is approximately $0.236\;dm^3$