Question

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in

the following composition reaction?
____ N2 + ____ O2 -> ____ NO

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with
NaOH how many moles of NaCl will be formed?
____ HCl + ____ NaOH->

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and
burn 150 moles of octane how many moles of carbon dioxide are you producing?

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how
many moles of magnesium oxide will be formed?
____ Mg + ____ O2 -> ____ MgO

5. . Lithium reacts with water in a single replacement reaction. How many moles of
hydrogen gas a produced by 10 grams of lithium?
____ Li + ____ HOH ->

6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How
many grams of barium chloride are required to react with 5 moles of sodium sulfate?

7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of
magnesium carbonate are heated?
____ MgCO3 -> ____ MgO + ____ CO2

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of
gold will be produced?
___ AuCl3 -->

9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

Answer 1

1. 10 moles of NO

2. 25 moles of NaCl

3. 1200 moles CO₂

4. 1.042 mole of MgO

5. 0.714 moles H₂ gas

6. 1040 g of BaCl₂

7. 9.5 g

8. 45.44 g of Au

9. 15 g of AlCl₃

Step-by-step explanation:

Ans 1.

Data Given:

moles of Oxygen = 5 moles

moles of nitrogen monoxide = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

N₂ + O₂ -----------> 2NO

1 mol 2 mol

So if we look at the reaction 1 mole Oxygen (O₂) gives 2 moles of nitrogen monoxide NO, then how many moles of nitrogen monoxide will be produced by 5 moles of Oxygen (O₂)

For this apply unity formula

1 mole of O₂ ≅ 2 moles of NO

5 mole of O₂ ≅ X moles of NO

By Doing cross multiplication

moles of NO = 2 moles x 5 moles / 1 mole

moles of NO = 10 mole

5 mole of O₂ will produce 10 moles of NO

________________

Ans 2.

Data Given:

moles of HCl = 25 moles

moles of NaCl = ?

Solution:

To solve this problem we have to look at the reaction

Neutralization Reaction:

HCl + NaOH -----------> NaCl + H₂O

1 mol 1 mol

So if we look at the reaction 1 mole HCl gives 1 moles of NaCl, then how many moles of NaCl will be formed by 25 moles of HCl

For this apply unity formula

1 mole of HCl ≅ 1 moles of NaCl

25 mole of HCl ≅ X moles of NaCl

By Doing cross multiplication

moles of NaCl = 1 moles x 25 moles / 1 mole

moles of NaCl = 25 mole

25 mole of HCL will form 25 moles of NaCl

________________

Ans 3.

Data Given:

moles of C₈H₁₈ = 150 moles

moles of CO₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2C₈H₁₈ + 25O₂ -----------> 16CO₂ + 18H₂O

2 mol 16 mol

So if we look at the reaction 2 mole C₈H₁₈ gives 16 moles of CO₂, then how many moles of CO₂ will be Produce by 150 moles of C₈H₁₈

For this apply unity formula

2 mole of C₈H₁₈ ≅ 16 moles of CO₂

150 mole of C₈H₁₈ ≅ X moles of CO₂

By Doing cross multiplication

moles of CO₂= 16 moles x 150 moles / 2 mole

moles of CO₂ = 1200 mole

150 mole of C₈H₁₈ will form 1200 moles of CO₂

______________________

Ans 4.

Data Given:

mass of Mg = 25 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Mg + O₂ -----------> 2MgO

2 mol 2 mol

Convert moles of Mg to mass

Molar mass of Mg = 24 g/mol

So,

2Mg + O₂ -----------> 2MgO

2 mol (24 g/mol) 2 mol

48 g 2 mol

So if we look at the reaction 48 g of Mg gives 2 moles of MgO, then how many moles of MgO will be Produce by 25 g of Mg

For this apply unity formula

48 g of Mg ≅ 2 moles of MgO

25 g of Mg ≅ X moles of MgO

By Doing cross multiplication

moles of MgO = 2 moles x 25 g / 48 g

moles of MgO = 1.042 mole

25 g of Mg will form 1.042 moles of MgO

______________________

Ans 5.

Data Given:

mass of Li = 10 g

moles of H₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Li + 2 H₂O -----------> 2LiOH + H₂

2 mol 1 mol

Convert moles of Li to mass

Molar mass of Li = 7 g/mol

So,

2Li + 2H₂O -----------> 2LiOH + H₂

2 mol (7 g/mol) 1 mol

14 g 1 mol

So if we look at the reaction 14 g of Li gives 1 moles of H₂ gas, then how many moles of H₂ gas will be Produce by 10 g of Li

For this apply unity formula

14 g of Li ≅ 1 moles of H₂

10 g of Li ≅ X moles of H₂

By Doing cross multiplication

moles of H₂ = 1 moles x 10 g / 14 g

moles of H₂ = 0.714 mole

10 g of Li will form 0.714 moles of H₂

______________________

Ans 6.

Data Given:

moles of Na₂SO₄= 5 moles

mass of BaCl₂ = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

Na₂SO₄ + BaCl₂ -----------> 2NaCl + BaSO₄

1 mol 1 mol

Convert moles of BaCl₂ to mass

Molar mass of BaCl₂ = 208 g/mol

So,

Na₂SO₄ + BaCl₂ -----------> 2NaCl + BaSO₄

1 mol 1 mol (208 g/mol)

1 mol 208 g

So if we look at the reaction 1 mole of Na₂SO₄ react with 208 g of BaCl₂, Then how many grams of BaCl₂ will react with 5 moles of Na₂SO₄

For this apply unity formula

1 mole of Na₂SO₄ ≅ 208 g of BaCl₂

5 mole of Na₂SO₄≅ X g of BaCl₂

By Doing cross multiplication

mass of BaCl₂ = 208 g x 5 moles / 1 mole

mass of BaCl₂ = 1040 g

5 moles of Na₂SO₄ will react with 1040 g of BaCl₂

______________________

Ans 7.

Data Given:

mass of MgCO₃ = 20 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

MgCO₃ -----------> MgO + CO₂

1 mol 1 mol

Convert moles to mass

Molar mass of MgCO₃ = 84 g/mol

Molar mass of MgO = 40 g/mol

So,

MgCO₃ -------------> MgO + CO₂

1 mol (84 g/mol) 1 mol (40 g/mol)

84 g 40 g

So if we look at the reaction 84 g of MgCO₃ gives 40 g of MgO, then how many g of MgO will be Produce by 20 g of MgCO₃

For this apply unity formula

84 g of MgCO₃ ≅ 40 g of MgO

20 g of MgCO₃ ≅ X g of MgO

By Doing cross multiplication

mass of MgO = 40 g x 20 g / 84g

mass of MgO = 9.5 g

20 g of MgCO₃ will produce 9.5 g of MgO

________________

The remaining portion is in attachment