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

A pertinent reaction between Silver(I) nitrate AgNO₃ & barium bromide BaBr₂ is,

2 AgNO₃ (aq) + BaBr₂ (aq) 2 AgBr (s) + Ba(NO₃)₂ (aq)

Mole relationship is hence,

2 moles of AgNO₃ (aq)    1 mole of BaBr₂ (aq) 2 moles of AgBr (s) 1 mole of Ba(NO₃)₂ (aq)

Molar masses are as,

AgNO₃ (169.87 g/mol) , BaBr₂ (297.14 g/mol) , AgBr (187.77 g/mol) ,  Ba(NO₃)₂ (261.34 g/mol)

Mass relationship is hence,

339.74 g of AgNO₃ (aq)    297.14 g of of BaBr₂ (aq) 375.54 g of of AgBr (s) 261.34 g of Ba(NO₃)₂ (aq) ...(1)

Given:

Mass of  AgNO₃ = 73.5 g

Mass of BaBr₂ (aq) = 92.6 g

a) Deciding Limiting reagent.

By eq.(1)

If,   339.74 g of AgNO₃ (aq)    297.14 g of of BaBr₂ (aq)

Then, 73.5 g of AgNO₃ (aq) say "A" g of of BaBr₂ (aq)

On cross multiplication,

A * 339.74 g = 73.5 g * 297.14 g

A = 73.5 g * 297.14 g / 339.74 g

A = 64.28 g

I.e. Mass of BaBr2 needed for 73.5 g AgNO3 to react completely is just 64.28 g. We are given with 92.6 g of BaBr2.

Hence, AgNO3 is limiting reagent and yield of product is decided by Mass of AgNO3 used.

Mass of excess reagent unreacted = Mass of BaBr2 - Mass of Babr2 actually reacted = 92.60 - 64.28 = 28.32g.

b) Mass of products formed.

i) Mass of AgBr = ?

By Eq.(1) Assuming 100% reaction completion.

If, 339.74 g of AgNO₃ (aq) 375.54 g of of AgBr (s).

Then, 73.5 g of AgNO₃ (aq) say "B" g of of AgBr (s).

On cross multiplication,

B * 339.74 g = 73.5 g * 375.54 g

B = 73.5 g * 375.54 g / 339.74 g

B = 81.25 g

Mass of AgBr formed = 81.25 g

ii) Mass of Ba(NO₃)₂ = ?

By eq.(1), assuming 100 % reaction completion,

If, 339.74 g of AgNO₃ (aq) 261.34 g of Ba(NO₃)₂ (aq)

Then,  73.5 g of AgNO₃ (aq) "C" g of Ba(NO₃)₂ (aq)

On cross multiplication,

C * 339.74 g = 261.34 g * 73.5 g

C = 261.34 g * 73.5 g / 339.74 g

C = 56.54 g

Mass of Ba(NO3)2 = 56.54 g.

Total Mass of product = Mass of AgBr + Mass of Ba(NO3)2 = 81.25 g + 56.54 g = 137.79 g

d) Mass of Reactants before reaction = Mass of AgNO3 + Mass of BaBr2 = 92.60 + 73.50 = 166.10 g.

Mass of reaction mixture after reaction = Mass of Ba(NO3)2 product + Mass of AgBr product + Mass of Unreacted BaBr2

= 81.25 g + 56.54 g + 28.32

= 166.1 g

Yes, Law of Conservation of mass is obeyed.

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