##### College general chemistry homework help

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How many grams of potassium chlorate must be decomposed to generate 30.0 g of oxygen?

What mass of oxygen is needed to oxidize 125 g of iron to iron(III) oxide?

Some cars can use butane (C4H10) as fuel. How many kilograms of oxygen are needed for the complete combustion of each kilogram of butane? Hint: the setup begins with one kg of butane.

What mass of anhydrous copper(II) sulfate (CuSO4) can be obtained by heating 125 g of copper(II) sulfate pentahydrate (CuSO45H2O)? Hint: the other product is water.

Calculate the mass percent of potassium nitrate in each of the following solutions: a) 2.51 g of KNO3 in 47.49 g of water

b) 4.22 kg of KNO3 in 51.43 kg of water

It is desired to prepare exactly 100.00 mL of sodium chloride solution. If 2.71 g NaCl is weighed out, transferred to volumetric flask, and water added to the 100-mL mark, what is the molarity of the resulting solution?

What volume (in mL) of 0.1008 M silver nitrate would be required to react completely with 0.0541 g or copper metal? Note: the copper product contains the copper(II) ion.

What is the percent yield of the reaction between 5.2 mL of 0.1233 M barium nitrate and 4.9 mL of 0.1256 M sodium sulfate if 0.140 grams of the precipitate are produced?

Acetylene (C2H2) burns in oxygen for form carbon dioxide and water. How many liters of carbon dioxide (at STP) are formed when 25.0 L of acetylene is burned at STP?

Acetylene (C2H2) burns in oxygen for form carbon dioxide and water. How many moles of carbon dioxide can be produced when 15.0 L of oxygen at STP reacts with 25.0 grams of acetylene?

I will give some answer because of shortage of time:-

**How many grams of potassium chlorate must be decomposed to generate 30.0 g of oxygen?**

2 KClO3 -> KCL + 3O2

Molar weight of O2 = 32 grams/mole (so close it doesn't matter)

30 grams/32grams/mole = 0.9375 moles

Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?)

now if we have 3 moles of O2 then we have 2 moles of KCl.

If we have one mole of O2 then we have 2/3 moles of KCL

What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl

So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl

So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl

**What mass of oxygen is needed to oxidize 125 g of iron to iron(III) oxide?**

4Fe(s) + 3O₂ --> 2Fe₂O₃

From the balanced equation, we see that:

4 mol Fe = 2 mol Fe₂O₃ => 2 ol Fe = 1 mol Fe₂O₃.

In moles, the weight of iron (III) is 3632/55.85 = 65.03 mol.

Finally, from the mole ratio, we see that:

65 mol Fe = (65 mol Fe) * (1 mol Fe₂O₃)/(2 mol Fe)

==> 65 mol Fe = 32.5 mol Fe₂O₃.

Finally, we see that the atomic weight of Fe₂O₃ is:

Molar mass of Fe₂O₃

= 2 * Molar mass of Fe + 3 * Molar mass of O

= 2 * 55.85 g/mol + 3 * 16/g mol

= 159.7 g/mol.

Then, 32.5 mol of Fe₂O₃ weighs 32.5*159.7 = 5.19 x 10^3 g.

So 5.19 x 10^3 g of iron(III) oxide can be created.

Answer of the both question is given:-

**Acetylene (C2H2) burns in oxygen for form carbon dioxide and water. How many liters of carbon dioxide (at STP) are formed when 25.0 L of acetylene is burned at STP?**

**Acetylene (C2H2) burns in oxygen for form carbon dioxide and water. How many moles of carbon dioxide can be produced when 15.0 L of oxygen at STP reacts with 25.0 grams of acetylene?**

Balance the equation below

C2H2 + O2 → CO2 + H2O, need 2 CO2 to balance the C’s

C2H2 + O2 → 2 CO2 + H2O, need 2.5 O2’s to balance the O’s

C2H2 + 2.5 O2 → 2 CO2 + H2O, double all the coefficients to have whole numbers

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

According to the balanced equation above, 2 moles of C2H2 react with 5 moles of O2 to produce 4 moles of CO2 and 2 moles of H2O.

At STP, 1 mole of a gas has a volume of 22.4 liters

25.0 L of acetylene(at STP) = 25/22.4 moles of C2H2

Ratio of O2/C2H2 = 5 / 2 = 2.5

So you need 2.5 * (25/22.4) moles of O2 to completely react with 25 liters of C2H2

Liters of O2 = moles of O2 * 22.4 = 2.5 * 25/22.4 * 22.4 = 2.5 * 25 = 62.5 liters of O2

Did you notice that the 22.4 canceled? Since, one mole of any gas occupies 22.4 liters, the mole ratios are volume ratios.

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

At a constant temperature, 2 liters of C2H2 will react with 5 liters of O2 to produce 4 liters of CO2 and 2 liters of H2O.

The volume ratios are the same as the mole ratios.

When I was teaching, we built proportions with the reaction equation.

For gasses, you use the given volume and the coefficients.

25 ……….. x ………… y ……….. z

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

25/2 = x/5

25/2 = y/ 5

25/2 = z/2

For mass – volume problems, you use the given mass and the product of the mass of 1 mole of a solid or liquid times the coefficient. For gasses, the product of the coefficient times 22.4.

Let’s burn 25 grams of C2H2 at STP.How many liters of O2 are required

25 g ………. x …….…. y …...….. z

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

(2*26) … (5*22.4)

25/52 = x/112

x = 53.8 liters

You will probably be doing these type of problems soon.

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