IB Chemistry SL 1.4.3 Molar Mass Question Bank

Practise IB Chemistry SL 1.4.3 with questions calculating molar mass and converting between mass, amount and chemical formulae.

[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 amount, in mol, of manganese(IV) oxide added.

n(\mathrm{MnO_2})=\frac{2.67\,\mathrm{g}}{86.94\,\mathrm{g\,mol^{-1}}}=0.0307\,\mathrm{mol}

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1.4.3—Molar mass (M) question 1