Question

How many grams of silver, Ag, can be produced from the reaction of 100.0 mL of 3.0 M AgNO3 with an excess of Al metal?

Answer 1

The reaction is:
3AgNO3 + Al → Al(NO3)3 + 3Ag
The number of moles of AgNO3 is 3.0M*0.1L= 0.3 moles
It can be seen from the reaction that the AgNO3/Ag ratio is 1:1, so the number of moles of Ag is 0.3.
Mass of Ag is 0.3moles*108g/mole=32.4g