# Question ebbe0

Feb 19, 2015

The mass of the collected hydrogen gas is $\text{0.0648 g}$.

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

$C {a}_{\left(s\right)} + 2 {H}_{2} {O}_{\left(l\right)} \to C a {\left(O H\right)}_{2} \left(a q\right) + {H}_{2} \left(g\right)$

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.

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

${P}_{\text{total") = P_("hydrogen") + P_("water") => P_("hydrogen") = P_("total") - P_("water}}$

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

$P V = n R T \implies n = \frac{P V}{R T}$

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

$\text{0.0324 moles" * ("2.0 g")/("1 mole") = "0.0648 g}$ ${H}_{2}$