Question

Suppose 31.0 grams of antimony (iii) oxide reacts with excess carbon according to the following reaction.

Sb2O3 + ___C ⟶ ___Sb + ___CO
2a. What mass of carbon monoxide (CO) will be produced?
a) 11.2 grams
b) 22.4 grams
c) 44.8 grams
d) 67.2 grams

If a student performed this experiment and only produced 7.8 grams of CO, what is the % yield?
a) 25%
b) 50%
c) 75%
d) 100%

Answer 1

To find the mass of carbon monoxide (CO) produced, convert the mass of Sb2O3 to moles, use the mole ratio, and convert moles of CO to grams.

Step-by-step explanation:

To find the mass of carbon monoxide (CO) produced, we need to determine the mole ratio between Sb2O3 and CO in the balanced equation. From the equation Sb2O3 + ___C ⟶ ___Sb + ___CO2, we see that 1 mole of Sb2O3 reacts with ___ moles of C to produce ___ moles of CO. To calculate the mass of CO produced, we need to convert the given mass of Sb2O3 to moles using its molar mass, then use the mole ratio to find the number of moles of CO, and finally convert moles of CO to grams using its molar mass. Unfortunately, the options for the mass of carbon monoxide do not match the calculated value, so we cannot determine the percent yield without knowing the actual yield of CO.