How many moles of aluminum nitrate are obtained from the reaction of 0.75 mol of silver nitrate with a sufficient amount of aluminum?

Md Ariful Islam

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Md Ariful Islam

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Solution:

Balanced chemical equation:

Al + 3AgNO3 → Al(NO3)3 + 3Ag

According to the equation above: n[AgNO3] / 3 = n[Al(NO3)3]

Hence,

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.

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