# While you cook on your gas grill, 3.7 grams of CO_2 are produced. How many moles of propane, C_3H_8 are used, using the equation C_3H_8+5O_2 -> 3CO_2+4H_2O?

Dec 2, 2015

0.084 moles of ${C}_{3} {H}_{8}$

#### Explanation:

Given mass of $C {O}_{2}$ = 3.7g
Molar mass of $C {O}_{2}$ = 12+2(16)= 44g/mol
No of moles of $C {O}_{2}$ produced = $\left(\text{given mass")/("molar mass}\right)$ = $\frac{3.7}{44}$ =0.084mol

According to balanced equation;
3 moles of $C {O}_{2}$ requires = 1 mole of ${C}_{3} {H}_{8}$
Thus, 0.084 moles of $C {O}_{2}$ will require= $\frac{1}{3} \times 0.084$mol of ${C}_{3} {H}_{8}$
= 0.028mol of ${C}_{3} {H}_{8}$
So, 0.028mol of ${C}_{3} {H}_{8}$ are used to produce 3.7g of $C {O}_{2}$