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In the reaction $2H_2 + O_2 \to 2H_2O$, $4$ moles of $H_2$ are mixed with $1$ mole of $O_2$. The mass of water formed is:
A$72$ g, taking all the hydrogen as the limit here
B$36$ g, the simple half of the hydrogen total here
C$36$ g, since O$_2$ limits to $2$ mol of water at $18$ g/mol
D$18$ g, just one mole of water, ignoring stoichiometry
Answer & Solution
Correct answer: C. $36$ g, since O$_2$ limits to $2$ mol of water at $18$ g/mol
O$_2$ is limiting: 1 mol O$_2$ $\to$ 2 mol H$_2$O = 36 g.