Some two equation/Two unkown Problem Magnesium and aluminum react with acid to g

Question

Some two equation/Two unkown Problem

Magnesium and aluminum react with acid to give hydrogen gas. What volume of hydrogen gas measured at 25°C and a pressure of 0.985 atm can be obtained from 4.00 g of an alloy that is 20.0 Mg and 70.0 Al? Mg (s) 2 H (aq) H2 (g) Mg (aq) +2 2 Al (s) 6 H (aq) 3 H2 (g) 2 Al 3 (aq) 1. A 2.05 g sample of an iron-aluminum alloy (ferroaluminum) is dissolved in excess HCI (aq) to produce 0.105 g H2 (g). What is the percent of aluminum, by mass, of the ferroaluminum? Fe (s) 2 HCl (aq) FeCl2 (aq) H2 (g) 2 Al (s) 6 HCl (aq) 2 AICI3 (aq) 3 H2 (g)

Explanation / Answer

The balanced equation for the reaction-

Mg(s) + 2H+(aq)---> Mg^2+(aq)+H2(g) ------(1)

2Al(s) + 6H+(aq)---> 2Al^3+(aq)+3H2(g) ------ (2)

Mass of sample = 4g

20% Mg in sample

Mass of Mg = 4 x 20/100 = 0.8 g

70% Al in sample

Mass of Al = 4 x 70/100 = 2.8 g

Step -1 Calculate the moles of H2 produced in reaction 1-

Use the molar mass to convert grams of Mg to moles of Mg

.Molar mass of Mg = 24.31 g/mol

Moles of Mg = 0.8g/24.31g/mol = 0.033

Now calculate the moles of H2 produced –

The balanced equation tells us that 1 mol Mg gives 1 mol H2

So, 0.033 mol Mg x1 mol H2 /1mol Mg = 0.033 mol H2

Step -2 calculate the moles of H2 produced in reaction 2-

Use the molar mass to convert grams of Al to moles of Al

.Molar mass of Al = 26.98 g/mol

Moles of Al = 2.8g/26.98g/mol = 0.104

Now calculate the moles of H2 produced –

The balanced equation tells us that 2 moles Al gives 3 moles H2

So, 0.104 mol Al x3 mol H2 /2mol Al = 0.156 mol H2

Step - 3 find the total number of moles of H2 produced

0.033moles + 0.156 moles = 0.189 moles H2

Step -4 Calculate the volume of H2

P = 0.985 atm (given)

T = 250C = 298 K

R = 0.082 L atm K1 mol1

We can calculate the volume of H2 by using Ideal Gas Law:

PV = n RT

V = n RT/P

V = (0.189 moles x 0.082 L atm K1 mol1 x 298K)/0.985 atm

V = 4.688L

Volume of H2 = 4.69 L

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