Question

What mass of solid carbon should be used to actually produce 60.0 g carbon monoxide according to the equation that follows if the percent yield of the reaction is 78.5%? (HINT: balance the equation first)

SiO₂ + C -> SiC + CO

I know the answer is 49.2g, but I got 32.75g and I don't know what I did wrong, so if someone could show their steps that would be great.

Answer 1

actual mass = 60.0 g

use:

% yield = actual mass * 100 / theoretical mass

78.5 = 60.0*100 / theoretical mass

theoretical mass = 76.4 g

The balanced reaction is:

2 SiO2 + 6 C —> 2 SiC + 4 CO

Molar mass of CO = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass of CO = 76.4 g

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

= 76.4/28.01

= 2.7276 mol

From balanced chemical reaction, we see that

when 4 mol of CO reacts, 6 mol of C is formed

mol of C formed = (6/4)* moles of CO

= (6/4)*2.7276

= 4.0914 mol

Molar mass of C = 12.01 g/mol

mass of C = number of mol * molar mass

= 4.0914*12.01

= 49.1g

Answer: 49.1 g