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3 years ago

What is the approximate percent yield for this reaction? Ba(NO3)2 + Na2SO4 —> BaSO4 + 2 NaNO3

Chemistry

2 answers:

melomori [17]3 years ago

6 0

The percent yield for this reaction is 91%.

<u>Explanation</u>:

Ba(NO3)2 + Na2SO4 — > BaSO4 + 2NaNO3

Assume that 0.45 mol of Ba(NO3)2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4.

Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield.

Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)

= 96 g BaSO4

Beginning with 0.45 mol of Ba(NO3)2, compute the theoretical yield of BaSO4. Mole proportion of Ba(NO3)2 to BaSO4 is 1:1.

0.45 mol Ba(NO3)2 x (1 mol BaSO4/1 mol Ba(NO3)2) x (233.38 g BaSO4/1 mol BaSO4) = 105 g BaSO4

% Yield = (Actual Yield/Theoretical Yield) x 100%

= (96/105) x 100%

= 91%.

pantera1 [17]3 years ago

4 0

Answer:

Is (B)

Explanation:

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