A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess

Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H_2(g) is collected over water at 29 degree C and 752 torr, the volume is found to be 345 mL. The vapor pressure of water at 29 degree C is 30.0 torr. how many moles of H_2 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 (ie., 0.5000x). How many moles of H_2 can be produced for y grams Al in magnesium-aluminum alloy? The molar mass of Al is 26.96 g/mol. Express your answer in terms of y to four decimal places(i.e., 0.5000y).

Explanation / Answer

mass of Mg = x g

Atomic mass of Mg = 24.31 g/mol

Number of moles of Mg = x/24.31 g/mol = 0.0411x mol

Reaction of Mg alloy with HCl to produce H2=>

Mg + 2HCl ---------> MgCl2 + H2

HCl is in excess, so whole of Mg will react. 1 mol of Mg produces 1 mol of H2.

So, 0.0411 x mol of Mg will produce 0.0411x mol of H2

mass of Al = y g

Atomic mass of Al = 26.98 g/mol

Number of moles of Al = y/26.98 = 0.0371y mol

Reaction of Mg-Al alloy with HCl to produce H2=>

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

1 mol of Al produces 3/2 mol of H2.

So, 0.0371y mol of Al will produce H2 = 3/2(0.0371y) = 0.0556y

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