Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29 degrees Celsius and752 torr, the volume is found to be 301 mL. The vapor pressure of water at 29 degrees Celsius is 30.0 torr.

How many moles of H2 can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass of Mg is 24.31 g/mol.

Express your answer in terms of x to four decimal places (i.e., 0.5000x).

Answer 1

change 311ml to 301ml!!





P_H2 = 0.950 atm (Dalton's Law of partial pressures)

n=Pv/RT
= (0.950 atm)(0.311 L) / (0.08206 (L*atm)/(mol*K))(302.15 K)
n_H2 = 0.011915983 mol

Balanced equations:

Al + 3HCl --> 3/2H2 + AlCl3
Mg + 2HCl --> H2 + MgCl2

By these equations, we know that every mole of Al will give us 1.5 moles of H2, and every mole of Mg will give 1 mole of H2.
We can therefore set up an equation for the mass of Al like this:
*Let a = the mass of MAGNESIUM*
Al = 0.250 g - a

With this equation in mind, we can setup two equations solving for 'n' of each element by dividing by its molar mass and multiplying by the molarratio:
n_Mg = a / 24.30 (1:1 ratio, so we don't have to multiply) --> number of moles of H2 produced by the reaction of Mg (now written as n_H') = a /24.30
n_Al = (0.250 g - a) / (26.98 g/mol) Because of the molar ratio shown above, we must multiply n_Al by 1.5 in order to get n_H2 produced by the reaction ofaluminum, hereafter known as n_H2"

Since we know the number of moles produced by the sum of the reactions, we can add these equations together and solve for n_H2.
(**note that your value will be different because you have a different volume**)
Set up the equation like this:

n_H2' + n_H2" = n_H2 = 0.011915983 mol

Sub in your individual equations for n_H2' and n_H2":

(a/24.3) + 1.5[(0.250-a)/26.98] = 0.011915983 mol

Rearrange and solve for a (mass of MAGNESIUM):
(26.98a + 9.1125 - 36.45a) / (24.3)(26.98) = 0.011915983
0.011915983 = 9.47a

a = 0.137298281 g

Once you have your 'a' value, divide it by the total mass (0.250 g) and multiply by 100%. This gives you the percentage of Mg.

(0.137298281 g / 0.250 g) * 100% = 54.9193 %

Since you want ALUMINUM, you must subtract the percentage of Mg from 100.

100 - 54.9193 = 45.08%

So, the mass percentage of aluminum is 45.08%.


I hope this is helpful!