Question

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`?

Answer 1

0.084 moles of `C_3H_8`

Explanation:

Given mass of `CO_2` = 3.7g
Molar mass of `CO_2` = 12+2(16)= 44g/mol
No of moles of `CO_2` produced = `("given mass")/("molar mass")` = `3.7/44` =0.084mol

According to balanced equation;
3 moles of `CO_2` requires = 1 mole of `C_3H_8`
Thus, 0.084 moles of `CO_2` will require= `1/3 xx 0.084`mol of `C_3H_8`
= 0.028mol of `C_3H_8`
So, 0.028mol of `C_3H_8` are used to produce 3.7g of `CO_2`