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 degrees Celsius and 752 torr, volume found to be 325mL. Vapor pressure of water at 29 Celsius is 30.0 torr. 0.0120 mol 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 answer in terms of x 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 answer in terms of y

Explanation / Answer

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 molar ratio: 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 of aluminum, hereafter known as n_H2"

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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