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.
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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