When a solution containing 320 g of barium hydroxide was reacted with a solution containing 227.2 g

| January 14, 2021 | education

When a solution containing 320 g of barium hydroxide was reacted with a solution containing 227.2 g of sodium sulfate, 312 g of barium sulfate were obtained. Determine the yield of the reaction product from the theoretically possible.

Let’s find the amount of substance Ba (OH) 2.
n = m: M.
M (Ba (OH) 2) = 171 g / mol.
n = 320 g: 171 g / mol = 1.87 mol (excess).
Let’s find the amount of the substance Na2SO4.
M (Na2SO4) = 142 g / mol.
n = 227.2 g: 142 g / mol = 1.6 mol.
Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.
Ba (OH) 2 + Na2SO4 = BaSO4 ↓ + 2NaOH.
According to the reaction equation, there is 1 mol of BaSO4 for 1 mol of Na2SO4. Substances are in quantitative ratios 1: 1.
n (BaSO4) = n (Na2SO4) = 1.6 mol.
Let’s find the mass of BaSO4.
M (BaSO4) = 233 g / mol.
m = n × M.
m = 233 g / mol × 1.6 mol = 372.8 g (theory).
372.8 g – 100%,
312 g – x%,
X = (312 × 100%): 372.8 = 83.69%.
Answer: 83.69%.



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