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Answer 1 · 2015-02-19T10:36:13

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

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

$C a_{(s)} + 2 H_{2} O_{(l)} \to C a {(O H)}_{2} (a q) + H_{2} (g)$ $Ca_((s)) + 2H_2O_((l)) -> Ca(OH)_2(aq) + H_2(g)$

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.

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

$P_{total} = P_{hydrogen} + P_{water} \Rightarrow P_{hydrogen} = P_{total} - P_{water}$ $P_("total") = P_("hydrogen") + P_("water") => P_("hydrogen") = P_("total") - P_("water")$

In this case,

$P_{hydrogen} = 988 mmHg - 31.82 mmHg = 956.18 mmHg$ $P_("hydrogen") = "988 mmHg" - "31.82 mmHg" = "956.18 mmHg"$

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

$P V = n R T \Rightarrow n = \frac{P V}{R T}$ $PV = nRT => n = (PV)/(RT)$

$n = \frac{\frac{956.18}{760} atm \cdot 641 \cdot 10^{- 3} L}{0.082 \frac{atm \cdot L}{mol \cdot K} \cdot (273.15 + 30) K} = 0.0324 moles hydrogen$ $n = (956.18/760"atm" * 641 * 10^(-3)"L")/(0.082 ("atm" * "L")/("mol" * "K") * (273.15 + 30)"K") = "0.0324 moles hydrogen"$

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

$0.0324 moles \cdot \frac{2.0 g}{1 mole} = 0.0648 g$ $"0.0324 moles" * ("2.0 g")/("1 mole") = "0.0648 g"$ $H_{2}$ $H_2$

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