150.0 mL of 0.200 M AgNO3 solution is reacted with 100.0 mL of 0.450 M NaBr solution, and the following occurs, which produces a precipitate of silver bromide, AgBr (molar mass = 187.8 g/mol). AgNO3(aq) + NaBr(aq) --> AgBr(s) + NaNO3(aq) If 5.12 g of AgBr is obtained, what is the percent yield of the reaction? (A) 100% (B) 90.9% (C) 72.7% (D) 60.6%
150.0 mL of 0.200 M AgNO3 solution is reacted with 100.0 mL of 0.450 M NaBr solution, and the following occurs, which produces a precipitate of silver bromide, AgBr (molar mass = 187.8 g/mol).
AgNO3(aq) + NaBr(aq) --> AgBr(s) + NaNO3(aq)
If 5.12 g of AgBr is obtained, what is the percent yield of the reaction?
(A) 100%
(B) 90.9%
(C) 72.7%
(D) 60.6%
The balanced reaction taking place is given as,
=> AgNO3 (aq) + NaBr (aq) -----> AgBr (s) + NaNO3 (aq)
Given : Concentration of AgNO3 = 0.200 M
Concentration of NaBr = 0.450 M
Volume of NaBr solution = 100.0 mL = 0.100 L
Volume of AgNO3 solution taken = 150.0 mL = 0.150 L (since 1 L = 1000 mL)
And mass of AgBr produced experimentally = 5.12 g
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