Wieviel Gramm Barium werden verbracht? Und Wie groß ist das Volumen des entstehenden Gases bei Normbedingungen?

Hi.

under standard conditions, 1 mole of gas corresponds to a volume of 24 liters – you have even given that.

You know m(H2O) 3.6 g.

From your reaction equation, you can determine the stoichiometric ratio of barium and water. One mole of barium reacts with 2 moles of water, i.e. n(Ba)=1/2 n(H2 O).

In addition, we know the molar mass of water and barium: M(H2O) = 18 g/mol and M(Ba) = 137.3 g/mol.

You can now replace n with m/M in the equation above:

m(Ba)/M(Ba) = 1/2 * m(H2O)/M(H2O).

Now we want to know the mass of Ba, we know the molar mass of Ba, so we bring it to the right:

m(Ba) = 1/2 * m(H2O)/M(H2O) * M(Ba).

We now use the known values:

m(Ba) = 1/2 * (3.6 g)/(18 g/mol) * 137.3 g/mol = 13.7 g.

In order to calculate the volume of H2, we are going to do the same. The following shall apply:

n(H2) = 1/2 * n(H2O).

On the right we replace n again with m/M, as we know m(H2O) and M(H2O). We want to Volume Calculate of hydrogen, whereby the molar volume with Vm =V/n helps us (that the formula applies, you recognize at the unit of Vm). The formula is converted to n and obtained n =V/Vm. We have to do that because we know Vm and want to know V. We replace:

V(H2)/Vm = 1/2 * m(H2O)/M(H2O).

Again we take Vm to the right and get:

V(H2) = 1/2 * m(H2O)/M(H2O) * Vm

= 1/2 * (3.6 g) / (18 g/mol) * 24 L/mol

= 2.4 L.

LG