College General Chemistry Question..any help?

Question

Chemistry and Percent yield...can anyone help?
1.LiOH + CO2 → Li2CO3 + H2O

If the percent yield of the above reaction is found to be 92.52 %, how many grams of Li2CO3 are produced from the complete reaction of 88.02 g of CO2 ?

2.What is the percentage of sulfur in NiSO4 by mass ?

3. 65.45 % C
5.49 % H
29.06 % O

4.13 g of the compound fills a 179.0 ml bottle with a pressure of 2593.4 mm Hg at a temperature of 24.6oC. What is the molecular formula of this compound ?

4.
A certain compound was found to have a molar mass of 74.122 g. A 1.512 g sample of this compound was burned in pure oxygen gas. The measured amount of CO2 produced was 3.591 g, while that of H2O was 1.837 g. No other products were produced in the reaction. How many moles of carbon were in the original sample that underwent combustion ?

Answer 1

1. 2LiOH + CO2 → Li2CO3 + H2O
CO2 ==> 12.01078 + 2*15.99943 ≈ 44.00964 g/mol
Li2CO3 ==> 2*6.9412 + 12.01078 + 3*15.99943 ≈ 73.89147 g/mol
x/ 73.89147 = 88.02/44.00964
x = 88.02*73.89147/44.00964
92.52x/100 = 88.02*73.89147*92.52/4400.964
92.52x/100 ≈ 136.7299 g ≈ 136.7 g

2. NiSO4 ==> 58.69342 + 32.0655 + 4* 15.99943 = 154.75612
Ni 37.92640%
S 20.72002%
O 41.35392%

3. (4.13 g)(22.4 L/0.179 L)(760/2593.4)(297.72/293.12) ≈ 153.8338 g/mol

4. x + O2 → CO2 + H20

3.591/44.00964 ≈ 0.08159576 mol. CO2
1.837/18.015324 ≈ 0.1019687 mol. H2O
or 8 mol. CO2 to 10 mol. H2O
2C4H10 + 13O2 → 8CO2 + 10H20
0.08159576 mol. C were in the original sample

Answer 2

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