Question

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO3)·2H2O(s) â 3Na2CO3(s) + CO2(g) + 5H2O(g) When 1.00 metric ton (1.00 à 103 kg) of trona is decomposed, 0.650 metric ton of Na2CO3 is recovered. What is the percent yield of this reaction?

a. 92.4%
b. 72.1%
c. 65.0%
d. 48.1%
e. 35.0%

Answer 1

a. 92.4%

Step-by-step explanation:

Based on the reaction:

2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)

To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:

Moles of trona:

1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = 4424 moles

The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):

4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = 6636 moles of Na₂CO₃.

The obtained moles of Na₂CO₃:

0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = 6133 moles

The ratio of obtained moles / theoretical moles gives:

6133 moles / 6636 moles = 0.924 = 92.4%

I hope it helps!