Anytime amounts are given for BOTH reactants you know it is a limiting reagent (LR) problem and the first thing you must do is determine the LR.
mols (NH4)2CrO4 = M x L = ?
mols Cr(NO3)3 = M x L = ?
Using the coefficients in the balanced equation, convert mols EACH one into mols of the product. It is likely that the two values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that number is the LR.
Use the smaller value of mols for the product and convert that to grams. g = mols x molar mass. This the theoretical yield (that is what you would get at 100%).
Then since the % yield is only 88.0%, multily the grams by 0.88. That is the grams of product you will actually receive.
In a 1L beaker, 203mL of 0.307 M ammonium chromate was mixed with 137mL of 0.269 M chromium(iii) nitrite.
3(NH4)2CrO4 + 2Cr(NO2)3 -> Cr2(CrO4)3 + 6NH4NO2
If the percent yield of the reaction was 88.0 %, what mass of solid product was isolated?
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1 answer