Questions: Calculate the mass of sodium chlorate that is used to form 7.5 g of oxygen given the balanced chemical equation. (The molar mass of O2 is 32.00 g / mol; NaClO3 is 106.44 g / mol.) 2 NaClO3(s) -> 2 NaCl(s) + 3 O2(g)

Solution Steps

First, we need to calculate the number of moles of oxygen (\(\mathrm{O}_2\)) produced from the given mass. The molar mass of \(\mathrm{O}_2\) is \(32.00 \, \mathrm{g/mol}\).

\[ \text{Moles of } \mathrm{O}_2 = \frac{7.5 \, \mathrm{g}}{32.00 \, \mathrm{g/mol}} = 0.2344 \, \mathrm{mol} \]

According to the balanced chemical equation:

\[ 2 \mathrm{NaClO}_3 \rightarrow 2 \mathrm{NaCl} + 3 \mathrm{O}_2 \]

The stoichiometric ratio between \(\mathrm{NaClO}_3\) and \(\mathrm{O}_2\) is \(2:3\). Therefore, the moles of \(\mathrm{NaClO}_3\) required are:

\[ \text{Moles of } \mathrm{NaClO}_3 = \frac{2}{3} \times 0.2344 \, \mathrm{mol} = 0.1563 \, \mathrm{mol} \]

Now, calculate the mass of \(\mathrm{NaClO}_3\) using its molar mass (\(106.44 \, \mathrm{g/mol}\)).

\[ \text{Mass of } \mathrm{NaClO}_3 = 0.1563 \, \mathrm{mol} \times 106.44 \, \mathrm{g/mol} = 16.63 \, \mathrm{g} \]

Final Answer