Question

1.

4.990 grams of C₂H₄ is reacted with 31.46 gram F₂ according to the following equation. How many gram of HF can be formed, assuming 100% yield?
C₂H₄ +   6F₂   -->   2CF₄ + 4 HF
Hint: Find limiting reagent.

		6.281 g
		8.830 g
		4.297 g
		17.21 g
		20.55 g
		18.34 g
		11.04 g
		25.91 g

2.

49.7 mL of a solution of NaOH is needed to neutralize a solution that contains 1.86 g of H₂SO₄  in water.    What is the molarity of the NaOH solution?

Hint--need to change g sulfuric acid to moles
  
H₂SO₄ + 2 NaOH   --> 2H₂O + Na₂SO₄

		0.763 M
		0.0404 M
		1.13 M
		0.481 M
		0.602  M
		2.23 M
		1.81  M
		2.94  M

3.

We find that 18.90 mL of a 2.50M KOH solution are required to titrate 35.0 mL of a H₂SO₄ solution. What is the molarity of the H₂SO₄ solution?
2KOH + H₂SO₄   --> K₂SO₄   + 2 H₂O

		0.675 M
		1.57 M
		0.0711  M
		0.128 M
		4.04 M
		0.903 M
		1.18 M
		0.0173 M

4.

How many mL of a 1.20 M  H₃PO₄ solution are required to titrate 85.5 mL of a 0.465 M LiOH   solution?
H₃PO₄(aq) + 3LiOH (aq) --> Li₃PO₄(aq) +   3H₂O(l).

		11.0 mL
		37.0 mL
		6.85 mL
		15.3 mL
		25.1 mL
		8.23 mL
		5.21 mL
		17.9 mL

5. In the following, what was reduced?  
2Na(s) + 2H₂O(l)   -->   2NaOH(aq) +   H₂(g)

		H2
		O in NaOH
		H in H2O
		Na
		Na in NaOH

Answer 1

Question 1

Balanced equation
C2H4 + 6 F2 ====> 2 CF4 + 4 HF

Mass of C2H4 = 4.99 gm

Moles of C2H4 = 4.99 g / 28.05 g/mol = 0.1778 Moles

Mass of F2 =  31.46 gm

Moles of F2 = 31.46 g / 37.99 g/mol = 0.8279 Moles

Limiting reagent is F2  

Moles of HF formed =  0.8279 Moles x 4/6 =  0.5519 Moles

Mass of HF produced =  0.5519 mol x 20.00 g/mol =  11.04 g

Option G is correct answer

Question 2  

H2SO4 + 2 NaOH ===> 2 H2O + Na2SO4  

Mass of H2SO4 = 1.86 gm

Moles of H2SO4 = 1.86 g / 98.047 g/mol = 0.01896 Moles

Moles of NaOH required =  0.03792 Moles

Volume of NaOH = 49.7 ml

Concentration of NaOH =  0.03792 x 1000 /49.7 = 0.7629 M

Option A is correct answer

Question 3

2 KOH + H2SO4 ===>  K2SO4 + 2 H2O

Volume of KOH = 18.9 ml

Concentration of KOH = 2.5 M

Moles of KOH = 18.9 x 2.5 / 1000  = 0.04725 Moles

Moles of H2SO4 required =  0.02362 Moles

Volume of H2SO4 = 35 ml

Concentration of H2SO4 =  0.02362 x 1000 /35 = 0.675 M

Option A is correct answer

Question 4

H3PO4(aq) + 3 LiOH(aq) ====> Li3PO4(aq) + 3 H2O(l)

Volume of LiOH = 85.5 ml

Concentration of LiOH = 0.465 M

Moles of LiOH = 85.5 x 0.465 / 1000  = 0.03975 Moles

Moles of H3PO4 required =  0.01325 Moles

Concentration of H3PO4 = 1.2  M

Volume of H3PO4 = 0.01325 x 1000 / 1.2 = 11.04 ml

Option A is correct answer

Question 5

H in H2O is reduced from +1 oxidation state to 0 oxidation state. Hence option C is correct answer