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# Analysis of a MagnesiumAluminum Alloy

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• Analysis of a MagnesiumAluminum Alloy
• ### Analysis of a MagnesiumAluminum Alloy

HilaryVar > 01-14-2020, 03:30 AM

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl (aq). When the liberated H2 is collected over water at 29 C and 752 torr, the volume is found to be 311 mL. The vapor pressure of water at 29 C is 30.0 torr. What is the mass percentage of aluminum in this alloy?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 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!Aluminum Magnesium AlloyThis Site Might Help You. RE: 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 H2 is collected over water at 29 C and 752 torr, the volume is found to be 311 mL. The vapor pressure of water at 29 C is 30.0 torr. What is the mass percentage of aluminum in this...I build aircraft,depends what aircraft it is. Some are made of a aluminum alloy mixed with other metal. Eurofighters outside skin is carbon fibre and contains many titanium parts. Then rivets hold everything together and they contain steel or alloyMg-Al + HCl -- MgCl2 + AlCl3 + H2 345ml gas collected over water vp water = 30torr total pressure - 30torr = pressure H2gas = 722torr = 0.95atm volume dry gas x 0.95atm = 345ml x 0.989atm volume dry gas = 359.3ml moles gas = pv/(rt) = 0.013moles 2Mg + 3Al + 10HCl -- 2MgCl2 + 2AlCl3 + 5H2 a) moles H2 = xgMg / 24.3g/mole x (5 H2 / 2Mg) = 0.103x b) moles H2 = ygAl / 27g/mole x (5H2 / 3Al) = 0.062yOther related question