# Predicting amounts of reaction products

What volume of $\ce{H2O(g)}$ is produced when 7.60 mol of $\ce{C2H4(g)}$ reacts at STP?

The reaction is $\ce{C2H4 + 3O2} = \ce{2CO2 + 2H2O}$.

Stoichiometry: 7.6 mol of $\ce{C2H4}$, for every 1 mol of $\ce{C2H4}$ there's 2 mol of $\ce{H2O}$ so $7.6\cdot2 = 15.2$ mol of $\ce{H2O}$. At STP 1 mol of any gas = 22.4 L, $15.2\cdot22.4 = 340.40$ L of $\ce{H2O}$.

Did I do it correctly, if not where did I go wrong and what method should I try instead of the one I did?

• Welcome to Chemistry Stack Exchange. Your working out looks correct to me and that should be the correct answer – Nanoputian Oct 3 '15 at 6:03
• Please note that, according to current IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $p=100\,000\ \mathrm{Pa}$. At this temperature and pressure, the molar volume of an ideal gas is not $V_\mathrm m=22.4\ \mathrm{l/mol}$; it actually is $V_\mathrm m=22.710\,947(13)\ \mathrm{l/mol}$. – Loong Oct 3 '15 at 9:53
• Please make titles more descriptive so that others can benefit from your question. – jerepierre Oct 3 '15 at 10:38