# How do you calculate the mass of water produced from the reaction of 24.0 g of H_2 and 160.0 g of CO_2? What is the limiting reagent?

Mar 14, 2017

$C {O}_{2} \left(g\right) + 4 {H}_{2} \left(g\right) \rightarrow C {H}_{4} \left(g\right) + 2 {H}_{2} O \left(g\right) + \Delta$
DeltaH = −165.0 *kJ*mol^-1

#### Explanation:

I assume this is the reaction $\text{(the Sabatier reaction)}$ to which you refer.

$\text{Moles of dihydrogen} = \frac{24.0 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1} = 11.9 \cdot m o l$

$\text{Moles of carbon dioxide} = \frac{160.0 \cdot g}{44.0 \cdot g \cdot m o {l}^{-} 1} = 3.64 \cdot m o l$

Clearly, ${H}_{2}$ is the reagent in deficiency, and according to this stoichiometry, $2.98 \cdot m o l$ carbon dioxide will react. And given this equivalence, $2 \times 2.98 \cdot m o l$ water will result, i.e. this represents a mass of $2 \times 2.98 \cdot m o l \times 18.0 \cdot g \cdot m o {l}^{-} 1 = 107.1 \cdot g$.

This reaction requires substantial catalysis, usually in the form of a ruthenium complex.