IB Chemistry SL 1.5 Stoichiometric Relationships Question Bank

Practise IB Chemistry SL 1.5 with evidence-led questions on stoichiometry, moles, equations, yields and quantitative analysis.

Covered syllabus points

1.5.1—Ideal gas model assumptions 1.5.2—Real gases 1.5.3—Molar volume 1.5.4—Ideal gas equation

[Maximum number: 4]

Alkanes are commonly occurring organic compounds.

(a)

The first four straight chain alkanes are gases at room temperature.

Boiling points of straight chain alkanes

[ 4 ]

(i)

Calculate the volume, in $\mathrm{dm}^{3}$ , occupied by 6.45 g of propane gas at 100 kPa and $15^{\circ} \mathrm{C}$ .

[ 2 ]

$\begin{aligned} \text { « } \mathrm{Mr}\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)=44.11 » \text { « } \mathrm{n}=\infty 0.146 \text { «mol» } \text { « } \mathrm{V}=\frac{0.146 \times 8.31 \times 288}{100} \Rightarrow 3.49 \text { «dm }{ }^{3} » \end{aligned}$

Award [2] for correct final answer.

Accept answers in the range 3.49-3.59 « $d m^{3}$ ».

(ii)

Outline why the volume occupied by propane(g) at very high pressure is higher than the value calculated using PV=nRT.

[ 2 ]

[Maximum number: 1]

Chlorine undergoes many reactions.

2.67 g of manganese(IV) oxide was added to $200.0 \mathrm{~cm}^{3}$ of $2.00 \mathrm{moldm} \mathrm{m}^{-3} \mathrm{HCl}$ .

\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{MnCl}_{2}(\mathrm{aq})

Calculate the volume of chlorine, in $\mathrm{dm}^{3}$ , produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

V(\mathrm{Cl_2})=0.0307\,\mathrm{mol}\times22.7\,\mathrm{dm^3\,mol^{-1}}=0.697\,\mathrm{dm^3}

[Maximum number: 3]

A hydrocarbon has the empirical formula $\mathrm{C}_{3} \mathrm{H}_{7}$ . When 1.17 g of the compound is heated to $85^{\circ} \mathrm{C}$ at a pressure of 101 kPa it occupies a volume of $400 \mathrm{~cm}^{3}$ .

Calculate the molar mass of the compound, showing your working.

$T=85+273=358\ \mathrm{K}$ $n=\frac{pV}{RT} =\frac{101\times0.400}{8.31\times358} =0.0136\ \mathrm{mol}$ $M=\frac{m}{n} =\frac{1.17}{0.0136} =86.2\ \mathrm{g\ mol^{-1}}$

Award [1 max] for correct final answer without working.

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Covered syllabus points

1.5 Ideal gases question 1

Question (a)

Question (i)

Question (ii)

1.5 Ideal gases question 2

1.5 Ideal gases question 3