Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

What volume, in liters, of 1.25-M nitric acid, HNO3is required to react with 65.0 g of tin?

What volume, in liters, of 1.25-M nitric acid, HNO3is required to react with 65.0 g of tin (HINT: Sn = 118.71 g/mol) according to the following equation?

Sn + 2 HNO3 ----> Sn(NO3)2 + H2

I know you start with converting Sn to moles but from there I'm lost. 

1 Answer

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  • 1 month ago

    Sn + 2 HNO3 → Sn(NO3)2 + H2

    (65.0 g Sn) / (118.71 g Sn/mol) x (2 mol HNO3 / 1 mol Sn) / (1.25 mol HNO3/L) = 0.876 L HNO3

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