Some two equation/Two unkown Problem
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 K−1 mol−1
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 K−1 mol−1 x 298K)/0.985 atm
V = 4.688L
Volume of H2 = 4.69 L
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Some two equation/Two unkown Problem Magnesium and aluminum react with acid to give hydrogen gas. What...
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