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 H2(g) is collected over water at 29 C and 752 torr, the volume is found to be 325 mL . The vapor pressure of water at 29 C 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).

How many moles of H2 can be produced from y grams of Al in magnesium-aluminum alloy? The molar mass of Al is 26.98 g/mol.

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

Explanation / Answer

No. of moles of H2 is 0.0132

The equations of reaction are:

Mg +2 HCl --> MgCl2 + H2
2 Al + 6 HCl --> 2 AlCl3 + 3 H2

Atomic mass of Mg is 24.31 g/mol and of Al = 26.98 g/mol.
If all of the alloy was Al, we have to calculate the mass of Al = 2/3*mol[H2]*26.98 g/mol = 237.4240 mg
If it all was Mg we calculate mol[H2]*24.31= 320.8920 mg
If Al is 85.5 % and Mg 14.5 % the alloy produces the amount of H2 given.

At first we have to calculate the gas volume at standard condition (25°C and 760 torr)

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