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3 years ago

What is the molar mass of 81.50g of gas exerting a pressure of 1.75atm on the walls of a 4.92L container at 307K?

2 answers:

7 0

M = 81.50g, mm = m/n
n = ???
PV = nRT --> n = PV/RT
n = (1.75)(4.92)/(.0821)(307)
n = 8.61/25.20 = .342
--> mm = m/n = 81.5/.342 = 238.58

algol [13]3 years ago

7 0

Answer: Molar mass of gas is 238.6 g/mol

Explanation:

Using ideal gas equation:

PV = nRT

P= pressure = 1.75 atm

V= volume = 4.92 L

n = no of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\farc{81.5}{M}$

R= gas constant =0.0821 Latm\molK

T = temperature = 307 K

R= gas constant = 8.314 J/Kmol

T= temperature = 190.7 K

M= molecular mass of gas = ?g/mol

$1.75\times 4.92=\frac{81.5}{M}\times 0.0821\times 307$

$M=238.6g/mol$

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4 years ago

Excess oxygen gas (O2) reacts with 244g of Iron (Fe) to produce 332 g of Fe2O3. What is the percent yield?

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Answer:

95.15%

Explanation:

To calculate the percent yield, we need the following formula:

$\% yield=\dfrac{actual yield}{theoreticalyield}\times100\%$

Solving for the theoretical yied, you need to predict how much of the product will be produced if we USE up the given.

Our given is 224g of Fe and we to get the theoretical yield, we need to figure out how much product will Fe produce supposing that we use up all the reactant.

First thing we do is get the balance equation of this chemical reaction:

4Fe + 3O₂ → 2Fe₂O₃

We get the ratio between Fe and the the product, Fe₂O₃

$\dfrac{4moles of Fe}{2molesofFe_{2}O_{3}}=\dfrac{2moles of Fe}{1moleofFe_{2}O_{3}}$

This basically means that we need 2 moles of Fe to produce 1 mole of Fe₂O₃. We'll use this later.

Now we let's use our given:

We need to first convert our given to moles. To do this, we need to determine how many grams there are of the reactant for every mole. We need to first get the atomic mass of the elements involved in the substance:

Iron(1)

Fe = 55.845(1) = 55.845g/mole(1)

Then we use this to convert grams to moles

$244g\times\dfrac{1mole}{55.845g}=4.369moles$

This means that there are 4.396moles of Fe in 244g of Fe.

This we will use to see how many moles of product we can produce given the moles of reactant by using the reactant:rproduct ratio.

$4.369moles of Fe\times\dfrac{1moleofFe_{2}O_{3}}{2molesofFe}=2.185 moles of Fe_{2}O_{3}$

So given 4.693 moles of Fe we can produce 2.185 moles of Fe₂O₃

The next step is to get how many grams of product there are given our calculation. We do this again by getting how many grams of Fe₂O₃ there are in 1 mole.

Fe(2) O(3)

Fe₂O₃=55.845(2) + 15.999(3)

= 111.69 + 47.997 =159.687g/mol

We then use this to solve for how many grams of product there are in 2.185 moles.

$2.185moles\times\dfrac{159.687g}{1mole}=348.92g$

This is our theoretical yield 348.92g of Fe₂O₃.

We can finally use our percent yield equation. Our actual yield is given by the probelm, 332g of Fe₂O₃ and we solved for our theoretical yield which is 348.92g of Fe₂O₃. We plug this in our formula and solve.

$\%yield=\dfrac{actual yield}{theoreticalyield}\times100\%$

$\%yield=\dfrac{332gofFe_{2}O_{3}}{348gofFe_{2}O_{3}}\times100\%=95.15\%$

So the answer is 95.15%

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