Question

How many grams of silver are formed when 2.0 g of tin react with 15 mL of 1.0 M silver nitrate solution?

Answer 1

Reaction of tin with silver nitrate is as follows

2Sn (s) +4 AgNO_​​​3 (aq)------------> 4 Ag (s)+2 Sn(NO_{​​​​​3}​)₂ (aq)

Atomic mass of tin : 118.71 g/mol

Number of moles of tin in 2 g of tin

= Weight of tin/Molar mass of Sn

= 2 g/118.7 g/mol = 0.0168 moles

Number of moles of silver nitrate in 15 mL of 1 M Silver nitrate solution is

= Molarity*Volume of solution in mL/1000

= 1.0 M * 15 mL/1000 = 0.015 moles

From above stoichiometric equation,

2 moles of tin reacts with 4 moles of silver nitrate

0.0168 moles of Sn requires ? moles of silver nitrate.

= 0.0168 moles * 4 moles / 2 moles = 0.0336 moles.

Similarly

4 moles of silver nitrate reacts with 2 moles of tin

0.015 moles of silver nitrate requires how many moles of Sn

= 0.015 moles * 2 moles/ 4 moles

= 0.0075 moles

Since 0.0168 moles of Tin has been taken, it is present in excess. So Silver nitrate is the limiting reagent and it decides that how much silver is formed.

4 moles of silver nitrate on reaction with excess of tin forms 4*108 g of silver is formed.

0.015 moles of silver nitrate on reaction with excess Tin forms? grams of Sn

= 0.015 moles * 4* 108 g /4 moles

= 1.62 g of Silver is formed.

Therefore

15 mL of 1.0 M Silver nitrate on reaction with 2 g of tin forms 1.62 g of Silver is formed.