# If 5.563 grams of CH_4 reacted with 25.00 grams of oxygen, how much CO_2 , in grams, can be produced? CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)?

Sep 24, 2016

Approx. thirty grams.

#### Explanation:

$\text{Moles of methane}$ $=$ $\frac{5.563 \cdot g}{16.043 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.3468 \cdot m o l$.

$\text{Moles of dioxygen}$ $=$ $\frac{25.00 \cdot g}{31.998 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.7813 \cdot m o l$.

Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

which requires 2 equiv dioxygen per equiv of methane. So we get 1 equiv of carbon dioxide, which has a mass of $2 \times 0.3468 \cdot m o l \times 44.010 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g