Question #7ddae

Jan 8, 2017

$\text{Option B}$

Explanation:

There were $\left(32 + 128\right) \cdot g = 160 \cdot g$ of reactant; there were $\left(88 + 72\right) \cdot g = 160 \cdot g$ of product. These data clearly illustrate the law of conservation of mass.

So a hydrocarbon was combusted, and the product mass, that of carbon dioxide and water, was equal to the reactant mass. Mass was conserved, as it is in every chemical reaction.

We can represent this by means of stoichiometrically balanced equation, which also conserves mass and charge:

$\text{CH"_4(g) + "2O"_2(g) rarr "CO"_2(g) + "2H"_2"O} \left(g\right)$

$\text{Moles of methane} = \frac{32.0 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 2.0 \cdot m o l$

$\text{Moles of dioxygen} = \frac{128.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 4.0 \cdot m o l$

$\text{Moles of carbon dioxide} = \frac{88.0 \cdot g}{44.0 \cdot g \cdot m o {l}^{-} 1} = 2.0 \cdot m o l$

$\text{Moles of water} = \frac{72.0 \cdot g}{18.0 \cdot g \cdot m o {l}^{-} 1} = 4.0 \cdot m o l$

And clearly the products and reactants were in stoichiometric proportion. Of course, I have done a lot of work for a multiple choice question. The simple fact is that mass is conserved for EVERY chemical reaction. You start with $n$ moles of stuff, you finish with an equivalent quantity.