Answer to Question #150121 in Chemistry for Cayden

Question

methanol is burned in a camp stove to provide heat for cooking if 18 g of methanol is burned what will the volume of oxygen measured at 37°C and 98.8 KPA required for complete combustion

Answer 1

2CH₃OH + 3O₂ = 2CO₂ + 4H₂O

n (O₂) = 3 x n (CH₃OH) /2

M (CH₃OH) = 32.04 g/mol

n (CH₃OH) = 18 / 32.04 = 0.56 mol

n (O₂) = 3 x 0.56 / 2 = 0.84 mol

PV=nRT

V=nRT/P

T = 37+273 = 310 K

R = 8.314 L⋅Pa⋅K⁻¹⋅mol⁻¹

V(O₂)=(0.84 x 8.314 x 310) / (98.8 x 1000) = 0.02 L