A $17.8 \cdot L$ $17.8L$ volume of propane gas for which $ρ = 493 \cdot g \cdot L^{- 1}$ $rho=493g*L^-1$ was completely combusted. What are (i) the molar quantities, and (ii) the mass of carbon dioxide emitted?

Question

2319 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T13:27:07

Approx. $26.4 \cdot k g \cdot C O_{2}$ $26.4*kg*CO_2$ are emitted.

We need (i) a stoichiometrically balanced equation:

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$ $C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(l)$

And (ii) the equivalent quantity of propane gas.

$Moles of propane$ $"Moles of propane"$ $=$ $=$ $(17.8*cancelLxx493*cancelg*cancel(L^-1))/(44.1*cancelg*mol^-1)=199*mol$

Now clearly, almost $600*mol$ of carbon dioxide will be evolved. Why clearly?

And thus mass of $"carbon dioxide"$ emitted,

$600*molxx44.01*g*mol^-1~=9*kg*CO_2$

A 17.8*L17.8⋅L17.8*L volume of propane gas for which rho=493*g*L^-1ρ=493⋅g⋅L−1rho=493*g*L^-1 was completely combusted. What are (i) the molar quantities, and (ii) the mass of carbon dioxide emitted?