Answer to Question #209687 in General Chemistry for mariam

Question

Methane, CH4, gas can be made in small quantities by the reaction shown below.

Al4C3(s) + 12 H2O (l)→ 4 Al(OH)3 (aq) + 3 CH4(g)

Determine the volume of methane gas that will be produced at SATP( 0 deg C and 100 kPa) when 1.73 grams of aluminum carbide, Al4C3 completely react with water? ( 6 marks) ( universal gas constant= 8.314 Kpa. L mol-1 K-1)

Answer 1

PV = nRT

P = 100 × 0.00987 = 0.987 atm

Ratio = 1:3

Molar Mass of methane = 16.04

Molar Mass of aluminum carbide = 143.96

143.96 ×1.73 = 249.1

= 1.73 × 16.04 × 3 = 83.25

= 4.73 moles

n = 4.73

T = 0 + 273 = 273

V = 4.73 × 8.314× 273/0.987

= 10877.19ml