Analysis of a Magnesium-Aluminum Alloy A 0.250-g sample of a magnesium-aluminum

Question

Analysis of a Magnesium-Aluminum Alloy 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 319 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 (i.e., 0.5000x). How many moles of H_2 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

A)

m = 0.25 g of Mg-Al alloy...

Pcorrected = (752-30)= 722 torr

PV = nRT,

n = PV/(RT) = 722*0.319/(62.4*302) = 0.012221 mol of H2

Assume

Mg + 2HCl = H2 + MgCl2

so,

0.012221 mol of H2 = 0.012221*2 = 0.024442 g of H2

0.012221 mol of H2 = 1/2*0.012221 = 0.0061105 mol of Mg present

mass = mol*MW = 0.0061105*24.3 = 0.1484 g of Mg will produce 0.024442 g of H2

so,

x = 0.024442/0.1484 = 0.16470 g of H2 will be produced by x g of Mg

B)

for Aluminium:

mass of Al = (0.25-0.1484) = 0.1016 g of Al

so

0.1016 g of Al = 0.024442 g of H2

so

y = 0.024442/0.1016 = 0.240570

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.