According to the equation

SnO2 + 2H2 -> Sn + 2H2O

What volume of H2, measured at 1 atm and 273 K, is required to react with 2.00 g of SnO2?

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`SnO_2 + 2H_2 rarr Sn + 2H_2O`

Molar mass of `SnO_2 = 150.7gmol`

Amount of `SnO_2` moles in 2g of `SnO_2 = 2/150.7=0.0133`

Mole ratio

`SnO_2:H_2 = 1:2`

Amount of` H_2` required `=0.0133xx2=0.0266`

Using `PV = nRT`

`P=1 atm`

`n=0.0266`

`R=0.08206 (atmL)/(molK)`

`T=273K`

`V=(nRT)/P`

`V=0.0266xx0.08206xx273/1=0.596L`

*So the volume of `H_2` required is 0.596L*

**Sources:**

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