Given the reaction 2C3H7OH + 9O2 -->CO2 + 8H2O, how many moles of oxygen are needed to react with 3.40 mol C3H7OH?

Oct 24, 2015

Given the balanced equation to follow, $\frac{9}{2}$ moles of dioxygen gas are required to combust 1 mol propanol. So the answer?

Explanation:

Equation:

${C}_{3} {H}_{7} O H + \frac{9}{2} {O}_{2}$ $\rightarrow$ $3 C {O}_{2} + 4 {H}_{2} O$

Is the equation above balanced? (If this was change from a large denomination bill I think you would immediately know whether you were being short-changed!). Let's see: LHS, $3 C$, $8 H$, and $10 O$; RHS, $3 C$, $8 H$, and $10 O$. So it is balanced.

Your starting conditions requires the combustion of 3.4 mol propanol. You do the arithmetic, but make sure that you are right.

Oct 27, 2015

Are you sure about the equation? seems this isn't balanced seeing that there are 5 carbons on the left side of the equation and only 1 carbon on the right side...

Explanation:

First, let's balance the above equation properly.

${C}_{3} {H}_{7} O H$ + ${O}_{2}$ = $C {O}_{2}$ + ${H}_{2} O$

left side:
C = 3
H = 8
O = 1 + 2 (DO NOT ADD IT UP YET)

right side:
C = 1
H = 2
O = 2 + 1 (DO NOT ADD IT UP YET)

balancing this equation we have

${C}_{3} {H}_{7} O H$ + $\textcolor{red}{\frac{9}{2}} {O}_{2}$ = $\textcolor{red}{3} C {O}_{2}$ + $\textcolor{red}{4} {H}_{2} O$

left side:
C = 3
H = 8
O = 1 + (2 x $\textcolor{red}{\frac{9}{2}}$) = 10

right side:
C = 1 x $\textcolor{red}{3}$ = 3
H = 2 x $\textcolor{red}{4}$ = 8
O = (2 x $\textcolor{red}{3}$) + (1 x $\textcolor{red}{4}$) = 10

Now that the equation is balanced, we can proceed in answering your initial question. Please note to ALWAYS check if the equation is balanced as your computation hangs on the correct coefficients in the chemical equation.

You are being asked to compute the number of ${O}_{2}$ moles needed to react with 3.40 moles of ${C}_{3} {H}_{7} O H$.

From the balanced equation, we know that for every 1 mole of ${C}_{3} {H}_{7} O H$, it needs $\frac{9}{2}$ (or 4.5) moles of ${O}_{2}$ to react. Thus,

$3.40$ $\cancel{\text{mol}}$ ${C}_{3} {H}_{7} O H$ x (4.5 "mol" O_2)/(1 cancel ("mol") C_3H_7OH) = $15.3$ $\text{mol}$ ${O}_{2}$

Therefore, the correct answer should be 15.3 moles of ${O}_{2}$