# According to the equationSnO2 + 2H2 -> Sn + 2H2OWhat 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*

Use stoichiometry:

`2.00"g(SnO_2)" = (1"mol(SnO_2)")/(150.71"g(SnO_2)") = (2"mol(H_2)")/(1"mol(SnO_2)") = 0.02654"mol(H_2)"`

Plug this value into the Ideal Gas Law and solve for volume:

`"PV" = "nRT"`

`(1"atm")"V" = (0.02654"mol")(0.08206"L"*"atm")(273"K")`

`"V" = 0.595 "L"`

In order to fully react with 2.00 g of SnO2, you will need **0.595 L of H2**. Note that you need three sig figs because you were given three sig figs in the problem.