chemistry review questions-calculations

Aya

Hi, I have some reviw questions for the upcomming chemistry exam, the problem is I don't have the answers so can some one read throught my work and check if it is wright? Thanks.
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1. An anaysis of a valitile liquid shoed that it wsa composed of 14.4%C, 2.37% H and 83.49% Cl. If 4.25 g of the liquid was vaporized and occupied 628 mL at 65 degrees celcuios and 112 kpa, what would be the molecular formula of the compund?
carbon
14.4% x 100g
= 14.4 g

m=14.4 g
n=1.199 mol
Mm=12.01 g/mol

hydrogen
2.37% x 100g
=2.37g

m=2.37 g
n=2.347mol
Mm=1.01 g/mol

Cholrine
83.49% x 100g
=83.49g

m=83.49 g
n=2.355mol
Mm=35.45 g/mol

so...

1.199 / 1.199 =1
2.37 / 1.1.99=2
2.355/ 1.199=2

Impirical Formula
C1 H2 Cl2

# mol of Impirical formula
PV=nrt
n=PV/rt
n=(112Kpa)(0.628L)/ 8.314kpaL/molL (388k)
n=0.025 mol

Molar mass of Impreical formula
Mm=m/n
Mm=4.25g/0.025 mol
Mm=170 g/mol

Molar mass of ( what is this the molar mass of???)
C1 H2 Cl2
84.93

Find a Factor

170g/mol/ 84.93 g/mol
=2

Molecular Formula
2(C1 H2 Cl2)

C2 H4 Cl4




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2. A gas has a volume of 40.0 ml at 40 degrees celcuios and 95 Kpa. What will be its volume at STP?

gas
40.0 mL = 0.04L
40 degrees celcioue=313 K
95 kPa=P

n=rt/pv
n= 8.134(313)/95(0.04)
n=27.3890 mol

v=nrt
v=27.3809(8.134)(273)/101.3
v=613.4948
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3. How many grams of sodium and required to product 2.24 L of Hydrogen gas, measured at 25 degrees celcuoiu adn 110 Kpa, according to the following reaction?

2Na + 2 H2O ---> 2 NaOH + H2O

n=rt/pv
n= 8.314(298)/110(2.24)
n=10.055

m=nMm
m=20.11 (22.99)
m=462.3289

B)calculate the volume of H2(g) produced when 54 g of Na(s) reacts with an excess of water at STP.

moles of Na
n=m/Mm
n=54/22.99
n=2.348

moles H
3.348/2
1.174

Volume
v=nrt/p
1.174(8.314)(273)/101.3
v=2.791

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4. An aquois solution has a volume of 2.0L and contains 36.0 g of glucose. If the molar mass of glucose is 180 g , what is the molarity of the solution?

molar mass of glucose
n=m/Mm
n=36.0/180
n=0.2

...what next?
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5. How would you prepare 100ml of 0.40 mol/L MgSO4 form a stock solution of 2.0 mol/L MgSO4?
n=0.100L/0.40 mol/L
0.25

v=n/c
v=0.25/2.0
v=0.125L
add 0.125L of MgSO4
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6. A solution contains 5.85 g of sodium chlride dissolved in 5.00 x 10^3 ml of water. What is the concentration of the sodium cholirde solution?
n=M/Mm
n=0.0568 mol


C=n/v
0.0568/5
=0.01136 mol/L
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7.what mass of potassium hydroxide is required to prepare 6.00 x 10^2 ml of a solution with a concentration of 0.225 mol/L
m=cv
0.225(0.5)
0.135mol

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8. What volume of 0.500mol/L sodium hyroxide solution can be prepared form 10.0 ml of a 6.00 mol/L solution?
C1V1=C2V2
6(0.01)=0.500(v2)
0.12=v2

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9. What is the mass of hydrochoric acid that is present in 500 mL of a solution containg 3.50 mol/L of HCl(aq)?
n=cv
3.50(0.5)
=1.75 mol

m=nMm
1.75(36.46)
63.805g

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10. Passing a park throught a mixture of hydrogen gas and oxygen gas produced water

a) calculate the mas of hydrogen needed to completly convert 4.00 g of oxygen into water

H2 + O = H2O

moles of O
n=m/Mm
4.00/18.025
0.22

moles of H
0.22 mol

mass of H
m=nMm
=0.22(1.01)
=0.224

b) calcute the number of moles of oxygen required to react with 12.5 moles of hydrogen gas
I dont know how to do this one
c)calculate the number of moles of water produced when 4.00 g of oxygen are used

I dont know how to do this one

***

Thanks for reading pleas post any corrections and solutions to the problems I don't know how to do. this is practise for an exam so pleas help!!!!

Hootenanny

HINT:

[tex]\text{M} = \frac{\text{moles}}{\text{volume}}[/tex]

You have almost already written the answer in your post, look again at the equation you written, i would just like to correct it however;

[tex]H_{2}_{(g)} + \frac{1}{2}O_{2}_{(g)} \rightarrow H_{2}O_{(l)}[/tex]

Now, all the answers you need are in the above equation.

Aya

4.
M=Mol/vol
M= 0.2mol/2.0L
M=0.1mol/L
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The teacher said we should only balance with whole nubbers so would it be

2H2 + O2 = 2H2O....?

moles of H
0.22 mol
sooo... moles of O is also 0.11mol?

c)calculate the number of moles of water produced when 4.00 g of oxygen are used

2H2 + O2 = 2H2O
2 : 1 = 2

oxygen
n=M/mm
n=4.00g/16.0
n=0.25mol

Water
0.25mol x 2
=0.5 mol

Are these ones right, and is everything elce right?

Thanks for your help!

Hootenanny

Spot on.
If your teacher restricts you to using whole numbers, then yes that is correct, however using half a diatomic molecules is acceptable.
In you original post you said, 12.5 mols of hydrogen.
Yes, that's spot on! I'm afraid I haven't checked the other as I haven't really got time, my apologies; but if you check back later I'm sure someone will have obliged.

Aya

^ Oh, ok thanks for all your help!