How many moles of aluminum nitrate are obtained from the reaction of 0.75 mol of silver nitrate with a sufficient amount of aluminum?
Balanced chemical equation:
Al + 3AgNO3 → Al(NO3)3 + 3Ag
According to the equation above: n[AgNO3] / 3 = n[Al(NO3)3]
Moles of Al(NO3)3 = Moles of AgNO3 / 3 = 0.75 mol / 3 = 0.25 mol
Moles of Al(NO3)3 = 0.25 mol
Answer: 0.25 mol of aluminum nitrate are obtained.