Question

C3H8(g)+5O2(g)→ 3CO2(g)+4H2O(l)

Answer 1

So we gots...${C}_{3} {H}_{8} + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

#### Explanation:

Clearly, four moles of water result from each mole of propane.

And if $1.638 \cdot g$ of propane are combusted this represents a molar quantity of...

$\frac{1.638 \cdot g}{44.10 \cdot g \cdot m o {l}^{-} 1} = 0.0371 \cdot m o l$...with respect to propane.

And thus .....

$\text{mass"_"carbon dioxide} = 3 \times 0.0371 \cdot m o l \times 44.01 \cdot g \cdot m o {l}^{-} 1 = 4.90 \cdot g$

$\text{mass"_"water} = 4 \times 0.0371 \cdot m o l \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 2.67 \cdot g$

How does the total mass of the products compare to the total mass of the reactants? Is this significant conceptually?

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