Question

What mass of water is produced from the complete combustion of 3.00×10−3g of methane?

Answer 1

The reaction has the following ratio of reactants to products:
\[CH_4+2O_2~~\implies~~2H_2O+CO_2\]
I'm a bit rusty on my chemistry, but if memory serves me correctly, you need to find the number of moles of each reactant and product so that the equation balances out.

What I would do is, using the given mass of methane, find its molar masses.
\[\textbf{Molecular masses:}\\
\begin{array}{c|c}
\text{Element}&\text{Mass }\left(\frac{\text{g}}{\text{mol}}\right)\\
\hline
C&\sim12\\
O&\sim16\\
H&\sim1\\
\hline
CH_4&\sim16\\
H_2O&\sim18\\
2O_2&\sim64\\
CO_2&\sim44\end{array}\]
Next convert to the number of moles of methane:
\[\begin{array}{c|r}
\text{Element}&\text{Number of moles}\\
\hline
CH_4&\left(3.00\times10^{-3}\text{g}\right)\left(\frac{1}{16}\frac{\text{mol}}{\text{g}}\right)=1.9\times10^{-4}\text{mol}
\end{array}\]
Since there is a ratio of 1 molecule of methane to 2 molecules of water, you know that for this reaction, \(3.8\times10^{-4}\text{mol}\) of water will be produced. Convert the number of moles to the mass:
\[\begin{array}{c|r}
\text{Element}&\text{Mass}\\
\hline
H_2O&\left(3.8\times10^{-4}\text{mol}\right)\left(18\frac{\text{g}}{\text{mol}}\right)=6.8\times10^{-3}\text{g}
\end{array}\]