Al(s) + HCl(aq) AlCl3(aq) + H2(g)
Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)
Consider the unbalanced equations above. A 0.200 g sample of a mixture of aluminum and zinc metal is reacted with an excess of hydrochloric acid. Both metals react with hydrochloric acid. The sample produces 168.2 mL of hydrogen gas at STP.
(a) What is the mass percent aluminum in the sample?
(b) What is the mass percent zinc in the sample?
Two equations in two unknowns means two equations solved simultaneously. You should first balance both equations (which you should have done on all of these before posting). Another note: you should find the arrow key on your computer. You don't know what's what if you can't tell the reactants from the products.
2Al(s) +6 HCl(aq)=> 2AlCl3(aq) + 3H2(g)
Zn(s) + 2HCl(aq)=> ZnCl2(aq) + H2(g)
Let X = mass Al metal.
and Y = mass Zn metal.
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eqn 1 is
X + Y = 0.200g
The second equation equates mols H2 with mols H2.
mols H2 from Zn is:
Y*(1/atomic mass Zn)*(1 mol H2/1 mol Zn)
mols H2 from Al is:
X(1 mol Al/atomic mass Al)*(3 mol H2/2 mol Al)
eqn 2 is
add mols H2 from X to mols H2 from Y and make it equal to 168.2/22,400
Solve for X and Y to obtain grams Al and Zn.
Then % Al = (grams Al/0.200)*100 = ?
%Zn = (grams Zn/0.200)*100 = ?
67.8
To find the mass percent of aluminum and zinc in the sample, we need to determine the amount of each metal that reacted and then calculate the mass percent using the formula:
Mass percent = (mass of element / total mass of sample) * 100
Let's calculate the amount of aluminum and zinc that reacted:
1. Determine the volume of hydrogen gas at STP:
168.2 mL of H2 gas
2. Convert the volume of gas to moles using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At STP, the pressure is 1 atm, and the temperature is 273.15 K.
Using the ideal gas law equation, we can calculate the number of moles of hydrogen gas produced:
n(H2) = (PV) / (RT)
n(H2) = (1 atm * 0.1682 L) / (0.0821 atm*L/mol*K * 273.15 K)
Note: We convert mL to L by dividing by 1000.
3. Calculate the number of moles of aluminum and zinc reacted:
Al(s) + 3HCl(aq) AlCl3(aq) + 1.5H2(g)
Zn(s) + 2HCl(aq) ZnCl2(aq) + 1H2(g)
By comparing the balanced equations, we can see that the ratio of moles of hydrogen gas to moles of aluminum and zinc is as follows:
For aluminum:
1.5 moles of H2 (from the balanced equation) reacts with 1 mole of aluminum.
So, for the given 0.1682 moles of H2, the moles of aluminum reacted will be:
n(Al) = 0.1682 moles * (1 mole Al / 1.5 moles H2)
For zinc:
1 mole of H2 (from the balanced equation) reacts with 1 mole of zinc.
So, for the given 0.1682 moles of H2, the moles of zinc reacted will be:
n(Zn) = 0.1682 moles * (1 mole Zn / 1 mole H2)
4. Calculate the mass of aluminum and zinc:
To determine the mass, we need the molar mass of aluminum and zinc. The atomic masses of aluminum (Al) and zinc (Zn) are 26.98 g/mol and 65.38 g/mol, respectively.
Mass(Al) = n(Al) * Molar mass(Al)
Mass(Zn) = n(Zn) * Molar mass(Zn)
5. Calculate the mass percent:
Mass percent Al = (Mass(Al) / Total mass of sample) * 100
Mass percent Zn = (Mass(Zn) / Total mass of sample) * 100
Substituting the calculated values, we can find the mass percent of aluminum and zinc in the sample.