Question

Silver(I) nitrate reacts with barium bromide to form silver(I) bromide and barium nitrate. If 73.5 g of silver(I) nitrate reacts with 92.6 g of barium bromide, how many g of barium nitrate will be formed? How many g of the excess reactant will be left after the reaction occurs? How many g of silver(I) bromide will be formed? Do your calculations verify the Law of Conservation of Matter?

Answer 1

2 AgNO3   +   BaBr2    ----------------> 2 AgBr + Ba(NO3)2

339.746 g      297.135 g                      375.54 g      261.34 g

73.5 g            92.6 g                                ??              ??

339.746 g AgNO3 ------------> 297.135 g BaBr2

73.5 g AgNO3   -----------------> ??

mass of BaBr2 required = 64.3 g

but we have 92.6 g . so this is excess .

limiting reagent is AgNO3.

mass of Ba(NO3)2 formed = 73.5 x 261.34 / 339.746

mass of Ba(NO3)2 formed = 56.5 g

mass of excess reactant left = 92.6 - 65.86

mass of excess reactant left = 28.3 g

mass of AgBr formed = 73.5 x 375.54 / 339.746

mass of AgBr formed = 81.2 g

yes, it satisfy Law of Conservation of Matter.

56.5 + 28.3 + 81.2 = 73.5 + 92.6

166 = 166