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When aqueous solutions of (NH4)2CrO4 and Ba(NO3 )2 are combined, BaCrO4 precipitates. Calculate the mass, in...

When aqueous solutions of (NH4)2CrO4 and Ba(NO3 )2 are combined, BaCrO4 precipitates. Calculate the mass, in grams, of the BaCrO4 produced when 3.5 mL of 0.164 M Ba(NO3 )2 and 4 mL of 0.742 M (NH4)2CrO4 are mixed. Calculate the mass to 3 significant figures.

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Answer #1

Ba(NO3)2(aq) + (NH4)2CrO4(aq) ------> BaCrO4(aq) + 2NH4NO3(aq)

moles of Ba(NO3)2 present = molarity*volume of solution in litres = 0.164*0.0035 = 0.000574

moles of (NH4)2CrO4 present = 0.742*0.004 = 0.002968

Clearly form the balanced reaction, Ba(NO3)2(aq) & (NH4)2CrO4(aq) reacts in the molar ratio of 1:1

Thus, Ba(NO3)2 is the limiting reagent

moles of (NH4)2CrO4 formed = moles of Ba(NO3)2 reacting = 0.000574

molar mass of (NH4)2CrO4 = 152.07 g/mole

Thus, mass of (NH4)2CrO4 formed = moles*molar mass = 0.0873 g

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