Who knows what to do?

Question

1. A hyrocarbon sample with a mass of 6 grams underwent combustion, producing 11 grams of carbon dioxide. If all of the carbon initally present in the compound was converted to carbon dioxide, what was the percent of carbon, by mass, in the hydrocarbon sample.

2. A sample of propane was completely burned in air at STP. The reaction occured as shown below
C3H8 + O2 --------> 3CO2 + 4H20
If 67.2 liters of CO2 were produced and all of the carbon in the CO2 came from the propane, what was the mass of the propane sample?

3. The concentration of sodium chloride in sea water is about 0.5 molar. How many grams of NaCl are present in 1kg of sea water?

4. What is the STP volume of SO2 gas produced when 145 grams of ZnSO3 are consumed?
ZnSo3(s) -----> ZnO(s) + SO2

All I really need to know is how to set up and solve these problems. Explain the process of solving these problems.

I really have a hard time with these stoichiometric problems because of my bad experiences with math in the past.

Answer 1

1. ?g C = 11g CO2 * 1mol CO2/44 g CO2 * 1mol C/1mol CO2 * 12g C/1mol C = 3g C
3g C/ 6 g hyrocarbon * 100 = 50%

2. ?g C3H8 = 67.2L CO2 * 1L C3H8 /3L CO2 * 1mol C3H8/22.4 L C3H8 * 44g C3H8/1mol C3H8 = 44 g C3H8

3. ?g H2O in 0.5M of NaCl solution = 1000g H2O - 0.5mol NaCl * 58.5g NaCl/1mol NaCl = 970.75g H2O
?g NaCl = 1000H2O * 29.25g Nacl/ 970.75g H2O = 30 g NaCl

4. ?L SO2 = 145g ZnSO3 * 1mol ZnSO3/145.5g ZnSO3 * 1mol SO2/1mol ZnSO3 * 22.4L SO2/1mol SO2 = 22.32L SO2

Answer 2

1) A dumb stoichiometry problem... I will see what I can do.

Mr of CO2 = 44
No. of mols = 0.25mols
Number of mols of Carbon = 0.25mols
0.25mols of carbon in 6g = 3g
Percentage of carbon = 50%

2) 1 dm3 = 1000cm3
67 200 cm3 = 67.2dm3
C3H8 + 5O2 --> 3CO2 + 4H2O
1 mol of C3H8 --> 3 mols of CO2
No. of mols of CO2 = 67.2/22.4
= 3mols
No. of mols of C3H8 = 1mol
Mass = 44g

3) Concentration = 0.5M/L
Mr of NaCl = 58.5
1kg of sea water = 1litre of sea water
number of mols of NaCl = 0.5mols
Mass = 58.5 x 0.5
= 29.25g

4) Mr of ZnSO3 = 145.3
No. of mols of ZnSO3 = 1.00mols
No. of mols of SO2 = 1.00mols
Volume = 1.00 x 22.4
= 22.4 dm3

Well... Thats what I can recall...

Edit: After reading STP, I found that I made some errors but nonetheless has changed it. Sorry. My answer and the following one is more or less the same except for question 3. I wonder how come there was an additional step...

How to prepare? I do not really get what you are trying to say... Are you going to like get propane and then react it with oxygen to get your final product? Sounds like a strange experiment...

As for the sea water, you mean taking the sea water and then distill it?

Make yourself clearer please.