Question

When you burn C4H10 gas in excess oxygen, the following reaction occurs: C4H10+

O2 CO2 + H₂O (This equation is not balanced, you must balance it first). How much O₂ in mass
will you need to burn 36.0 g of C4H10? Molar
mass (g/mol): C: 12, O: 16, H:1.

(A) 119.01 g
(B) 110.03 g
(C) 129.10 g
(D) 89.07 g

Answer 1

Step-by-step explanation:

Balance the equation to:

2 C4 H10 + 13 O2 =====> 8 CO2 + 10 H2O

The mole ratio of O2 to C4H10 is 13/2

36 g of C4 H2 = 36 / ( 4*12 + 1 *2) = .72 mole

so you will need 13/2 * .72 = 4.68 Moles of O2

4.68 moles * 32 gm/mole = 149.76 gm of O2 needed