Answer:
0.91g of Al
Explanation:
Al(NO3)3(aq) + 3Ag(s) -> Al + 3Ag(NO3)
We check and we see that the given reaction equation is balanced
Now, from the equation, 3 moles of Ag reacts to produce 1 mole of Al
Mole ratio is 3:1
Number of moles = mass/molar mass
Molar mass of Ag ~ 108g/mol
3 moles of Ag = m/108g/mol
m = 3 moles of Ag × 108g/mol
m= 324g
1 mole of Al = 27g
Therefore, 324g of Ag produces 27g of Al,
10.87g of Ag will produce [(10.87g×27g)/324g] of Al
= 0.91g of Al
