Answer to Question #153860 in Physical Chemistry for Yasir Nadeem

Question

How many grams of each product result from the following reac-

tions, and how many grams of which reactant are left over?

(a) 11.3 g NaCl2 + 13.5 g AgNO32 ¡1x g AgCl2 + 1y g NaNO32

(b) 12.65 g BaCl22 + 16.78 g H2SO42 ¡1x g BaSO42 + 1y g HCl2

Answer 1

A. NaCl + AgNO₃ ---> AgCl + NaNO₃

Molar mass of NaCl= 23+35.5= 58.5g

Molar mass of AgNO₃= 108 + 14 + 48 = 170g

Molar mass of AgCl= 108+35.5= 143.5g

Molar mass of NaNO₃= 23+14+48= 85g

58.5g of NaCl react with 170g of AgNO₃

11.3g of NaCl will react with 32.84g of AgNO₃

Since only 13.5g of AgNO₃ is present, it is the limiting reagent.

For AgCl:

170g of AgNO₃ yields 143.5g of AgCl

13.5g of AgNO₃will yield 143.5x13.5/170 = 11.4g of AgCl

For NaNO3:

170g of AgNO3 yields 85g of NaNO3

13.5g of AgNO3 will yield 85x13.5/170 = 6.75g of AgCl

Since 13.5g of AgNO3 reacted,

170g of AgNO3 reacts with 58.5g of NaCl

13.5g of AgNO3 will react with 58.5x13.5/170 = 4.65g of NaCl

Amount of NaCl left = 11.3-4.65= 6.65g of NaCl

B. BaCl2 + H2SO4 -----> BaSO4 + 2HCl

Molar masses:

BaCl2= 208.33g

H2SO4= 98g

BaSO4= 233.33g

HCl= 36.5g

98g of H2SO4 react with 208.33g of BaCl2

16.78g of H2SO4 will react with 35.67g

Since only 12.65g of BaCl2 ia present, it is the limiting reagent.

For BaSO4:

208.33g of BaCl2 yields 233.33g of BaSO4

12.65g of BaCl2 will yield 233.33x12.65/208.33 = 14.17g

For HCl:

208.33g of BaCl2 yields 2(36.5)g of HCl

12.65g of BaCl2 will yield 2(36.5)x12.65/208.33 = 4.43g of HCl

Since only 12.65g of BaCl2 is available.

208.33g of BaCl2 react with 98g of H2SO4

12.65g of BaCl2 will react with 98x12.65/208.33 = 5.95g

Amount of H2SO4 left = 16.78-5.95 = 10.83g