Question

Silicon dioxide reacts with powdered carbon at high temperatures to produce SiC and CO SiO2(s) + 3C(s) → SiC(s) + 2CO(g) When 4.16 g of SiO2 and 2.21 g C(s) are mixed together what mass of SiC will form if the percent yield is 74.5%? 2.07 g 1.83 g 1.95 g 2.78 g 2.46 g

Answer 1

SiO2 + 3C ----> SiC + 2CO

Molar mass of SiO2 = 60.08 g/mol

4.16 g / 60.08 g/mol => 0.0692 mol of SiO2

Molar mass of C = 12 g/mol

2.21 g / 12 g/mol => 0.1842 mol of C

1 equivalent of SiO2 requires 3 equivalent of C

So 0.0692 * 3 => 0.2076 mol of required C for completely react with SiO2

But available amount of C is 0.1842 mol

Hence 0.1842 / 3 => 0.0614 mol of SiO2 can react with available amount of C

C is limiting reactant.

Theoretical amount of SiC = 0.0614 mol

But actual yield is 74.5 % of theoretical yield.

Hence amount of actual yield = 0.0614 * (74.5/100) => 0.0457 mol

Molar mass of SiC = 40.11 g/mol

Amount of SiC = 0.0457 mol * 40.11 g/mol => 1.83 g of SiC

Answer 1.83 g