Question

Ch 4 Q 4.41 a The percentage yield of the following reaction is Chu cg) + 45 (g) -> CS2(g) + 2 H₂Sig) How many grams of sulfur would be needed constently 87%. to obtain 80.0g of CS2?

Answer 1

Actual mass of CS2 = 80.0 g

Use:

% yield = actual mass * 100 / theoretical mass

87 = 80.0 * 100 / theoretical mass

theoretical mass = 91.95 g

Molar mass of CS2,

MM = 1*MM(C) + 2*MM(S)

= 1*12.01 + 2*32.07

= 76.15 g/mol

mass of CS2 = 91.95 g

mol of CS2 = (mass)/(molar mass)

= 91.95/76.15

= 1.207 mol

According to balanced equation

mol of S reacted = (4/1)* moles of CS2

= (4/1)*1.207

= 4.83 mol

Molar mass of S = 32.07 g/mol

mass of S = number of mol * molar mass

= 4.83*32.07

= 1.549*10^2 g

Answer: 155 g