One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na₅(CO₃) ₂(HCO₃) $\rarr$ 2H₂O, as shown in the following reaction: 2Na₃(CO₃) (HCO₃) $\rarr$ 2H₂O(s) $\rarr$ 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
When 15 metric ton (1 *10³ kg) of trona is decomposed, 11 metric ton of Na₂CO₃ is recovered. What is the percent yield of this reaction? (1 metric ton = 10³ kg)

Question

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na₅(CO₃) ₂(HCO₃)

\rarr

2H₂O, as shown in the following reaction: 2Na₃(CO₃) (HCO₃)

\rarr

2H₂O(s)

\rarr

3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
When 15 metric ton (1 *10³ kg) of trona is decomposed, 11 metric ton of Na₂CO₃ is recovered. What is the percent yield of this reaction? (1 metric ton = 10³ kg)

Quizplus · Accepted Answer

Percent yield = (Actual yield/Theoretical yield) x 100Theoretical yield of Na₂CO₃ = 15 metric ton x (3 Na₂CO₃/2 Na₅(CO₃)₂(HCO₃)) x (106 g/1 mole) x (1 kg/1000 g) = 22 metric tonPercent yield = (11 metric ton/22 metric ton) x 100 = 50%

One Way of Obtaining Pure Sodium Carbonate Is Through the Decomposition