Question #ebbe0

1 Answer
Feb 19, 2015

The mass of the collected hydrogen gas is #"0.0648 g"#.

Once again, here's the reaction wou're working with

#Ca_((s)) + 2H_2O_((l)) -> Ca(OH)_2(aq) + H_2(g)#

The basic idea behind this reaction is that the hydrogen gas will be collected over water at a total pressure that includes the water vapor.

http://www.docbrown.info/page13/ChemicalTests/GasPreparation.htm

So, in order to determine the pressure of the hydrogen gas, you must remove the water vapor from the total pressure

#P_("total") = P_("hydrogen") + P_("water") => P_("hydrogen") = P_("total") - P_("water")#

In this case,

#P_("hydrogen") = "988 mmHg" - "31.82 mmHg" = "956.18 mmHg"#

Now just use the ideal gas law equation to solve for the moles of hydrogen produced

#PV = nRT => n = (PV)/(RT)#

#n = (956.18/760"atm" * 641 * 10^(-3)"L")/(0.082 ("atm" * "L")/("mol" * "K") * (273.15 + 30)"K") = "0.0324 moles hydrogen"#

Once again, use the units required for this value of R - atm, L, and K!

Now just use hydrogen's molar mass to determine the actual mass

#"0.0324 moles" * ("2.0 g")/("1 mole") = "0.0648 g"# #H_2#