A mixture contained zinc sulfide, ZnS, and lead sulfide, PbS. A sample of the mixture weighing 6.12 g was reacted with an excess of hydrochloric acid. The reactions are given below.

ZnS(s) + 2 HCl(aq) → ZnCl2(aq) + H2S(g)

PbS(s) + 2 HCl(aq) → PbCl2(aq) + H2S(g)
If the sample reacted completely and produced 1.038 L of hydrogen sulfide, H2S, at 23°C and 745 mmHg, what were the percentages of ZnS and PbS in the mixture?
ZnS

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To find the percentages of ZnS and PbS in the mixture, we need to calculate the moles of hydrogen sulfide (H2S) produced from the reactions and then use stoichiometry to determine the moles of ZnS and PbS.

Step 1: Calculate the moles of H2S
We know the volume (1.038 L) and the pressure (745 mmHg) of H2S. To convert these values to moles, we can use the ideal gas law equation:
PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature in Kelvin (23°C = 23+273 = 296 K)

First, convert the pressure from mmHg to atm:
745 mmHg ÷ 760 mmHg/atm = 0.980 atm

Now we can plug in the values into the ideal gas law equation:
0.980 atm × 1.038 L = n × 0.0821 L atm/mol K × 296 K

Simplifying the equation gives us:
n = (0.980 atm × 1.038 L) / (0.0821 L atm/mol K × 296 K)
n = 0.0415 mol

So, 0.0415 moles of H2S were produced.

Step 2: Determine the moles of ZnS and PbS
From the balanced equations, we can see that each mole of H2S is produced from one mole of ZnS or PbS.

Let's assume x moles of ZnS and y moles of PbS were in the original mixture.

Using the stoichiometry of the reaction, we can set up the following equation:
x + y = 0.0415 mol

Step 3: Calculate the percentages of ZnS and PbS
The percentages of ZnS and PbS in the mixture can be calculated using the following formulas:

Percentage of ZnS = (moles of ZnS / total moles of mixture) x 100%
Percentage of PbS = (moles of PbS / total moles of mixture) x 100%

To find the moles of ZnS and PbS, we can solve the system of equations:
x + y = 0.0415 (Equation 1)

We don't have enough information to solve for x and y directly. However, we know that ZnS and PbS are the only components in the mixture, so the sum of their percentages will be 100%.

Let's assume the Percentage of ZnS is given by p, then the Percentage of PbS will be (100 - p).

Substituting this assumption into Equation 1, we get:
x + (100 - p) = 0.0415

Simplifying the equation gives us:
x = 0.0415 - (100 - p)

Now we can calculate the moles of ZnS and PbS:
Moles of ZnS = 0.0415 - (100 - p)
Moles of PbS = 100 - p

Finally, we can calculate the percentages:
Percentage of ZnS = (moles of ZnS / total moles of mixture) x 100%
Percentage of ZnS = [(0.0415 - (100 - p)) / 0.0415] x 100%
Percentage of ZnS = [(p - 99.9585) / 0.0415] x 100%

Percentage of PbS = (moles of PbS / total moles of mixture) x 100%
Percentage of PbS = [(100 - p) / 0.0415] x 100%

Now you can substitute the appropriate value of p (based on the given range, usually from 0 to 100) to find the percentages of ZnS and PbS in the mixture.

Let X = mass ZnS

Let Y = mass PbS
MM = molar mass
=====================
You need two equations.
X + Y = 6.12 is one. The second is
X(MM H2S/MM ZnS) + Y(MMH2S/MMPbS) = grams H2S.
The g H2S must be calculated from PV = nRT and solve for n=number of moles H2S, then convert to grams, knowing 22.4 L H2S at STP will have a mass of 34.08 grams.

Solve the two equation simultaneously which will give you grams ZnS and grams PbS. Then percent ZnS = (mass ZnS/mass sample)*100 = ?? and
%PbS = (mass PbS/mass sample)*100 = ??