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Stoichiometric Calculations

We allow 32...

Question

We allow $32.0$ g of methane ( $C H_{4}$ ) to react as completely as possible with excess of $O_{2}$ to form $C O_{2}$ and $H_{2} O$ . How many moles of $O_{2}$ will react?

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Solution

The balanced chemical equation for the combustion reaction is,
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
Thus, 1 mole of methane reacts with 2 moles of oxygen.
2 moles of methane (32.0 g) will react with 4 moles of oxygen.

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