Question

a)What is the mass (in grams) of the solid product when 7.5 g of silver nitrate reacts in solution with excess sodium carbonate?AgNO₃(aq) + Na₂CO₃(aq) --> Ag₂CO₃(s) + NaNO₃(aq) [unbalanced]

b)What is the mass (in grams) of the precipitate when 7.62 g of silver nitrate reacts in solution with excess sodium phosphate?AgNO₃(aq) + Na₃PO₄(aq) --> Ag₃PO₄(s) + NaNO₃(aq) [unbalanced]

Answer 1

a)

Let; We have given;

7.5g of Silver nitrate;

Calculating the number of moles of Silver nitrate;

7.5g AgNO₃ x (1mole/169.87g) =0.04415 moles of AgNO₃

Now, The balanced reaction will be ;

2AgNO_3(aq) + Na₂CO_3(aq) Ag₂CO_3(s) + 2NaNO_3(aq)

Now; Calculating the moles of Ag₂CO₃ that will be formed from 0.02643 moles of AgNO₃

0.04415 moles of AgNO₃ x (1 mole of Ag₂CO₃/2 moles of AgNO₃) =0.02207 moles of Ag₂CO₃

Converting these moles of precipitate to grams;

0.02207moles of Ag₂CO₃ x( 275.74g/ 1mole) =6.087 g of Ag₂CO_3(s)

b)

Let; We have given;

7.62g of Silver nitrate;

Calculating the number of moles of Silver nitrate;

7.62g AgNO₃ x (1mole/169.87g) =0.04485 moles of AgNO₃

Now, The balanced reaction will be ;

3AgNO_3(aq) + Na₃PO_4(aq) Ag₃PO_4(s) + 3NaNO_3(aq)

Now; Calculating the moles of Ag₃PO₄ that will be formed from 0.04485 moles of AgNO₃

0.04485 moles of AgNO₃ x (1 mole of Ag₃PO₄/ 3 moles of AgNO₃) =0.01495 moles of Ag₃PO₄

Converting these moles of precipitate to grams;

0.01495 moles of Ag₃PO₄ x( 418.58g/ 1mole) =6.26 g of Ag₃PO_4(s)