Question

1. Sodium chlorate decomposes into sodium chloride and oxygen gas as seen in the equation below.

­­2NaClO3­ --> 2NaCl +3O2



How many grams of NaClO3­ were needed to produce 9 grams of O2? Round your answer to the nearest whole number.

Round your answer to the whole number. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element

Molar Mass

Sodium

23

Chlorine

35.5

Copper

63.5

Oxygen

16

2. S + 6 HNO3 --> H2SO4 + 6 NO2 + 2 H2O

In the above equation how many grams of water can be made when 2 grams of HNO3 are consumed?

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element

Molar Mass

Hydrogen

1

Nitrogen

14

Sulfur

32

Oxygen

16


3. 2 NH3 + 3 CuO g 3 Cu + N2 + 3 H2O

In the above equation how many grams of N2 can be made when 138 grams of CuO are consumed?

Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element

Molar Mass

Hydrogen

1

Nitrogen

14

Copper

63.5

Oxygen

16


4. For the reaction 2HNO3 + Mg(OH)2 → Mg(NO3) 2 + 2H2O, how many grams of magnesium nitrate are produced from 8 grams of nitric acid, HNO3 ?

Round your answer to the nearest tenth. Do not round your answers until the last step of the problem. If you do, the computer might mark your answer incorrect and you will be asked to complete your second trial for this assessment.

Use the following molar masses:

Element Molar Mass
Hydrogen 1
Magnesium 24
Nitrogen 14
Oxygen 16

Answer 1

1. Let's first find the moles of O2 produced:

9 g O2 * (1 mol O2 / 32 g O2) = 0.28125 mol O2

Now let's use the stoichiometry of the reaction to find the moles of NaClO3 needed:

0.28125 mol O2 * (2 mol NaClO3 / 3 mol O2) = 0.1875 mol NaClO3

Now, let's find the mass of NaClO3:

0.1875 mol NaClO3 * (23 g/mol Na + 35.5 g/mol Cl + 3 * 16 g/mol O) = 0.1875 mol NaClO3 * 106.5 g/mol NaClO3 ≈ 19.97 g NaClO3

Answer: 20.0 g NaClO3

• 2. Let's find the moles of HNO3 consumed:
• 2 g HNO3 * (1 mol HNO3 / (1 + 14 + 16*3) g HNO3) = 2 g HNO3 * (1 mol HNO3 / 63 g HNO3) ≈ 0.0317 mol HNO3
• Now let's use the stoichiometry of the reaction to find the moles of H2O produced:
• 0.0317 mol HNO3 * (2 mol H2O / 6 mol HNO3) ≈ 0.0106 mol H2O
• Now, let's find the mass of H2O:
• 0.0106 mol H2O * (2 * 1 + 16) g/mol H2O = 0.0106 mol H2O * 18 g/mol H2O ≈ 0.1908 g H2O

Answer: 0.1908 g H2O

3. Let's find the moles of CuO consumed:

138 g CuO * (1 mol CuO / (63.5 + 16) g CuO) = 138 g CuO * (1 mol CuO / 79.5 g CuO) ≈ 1.735 mol CuO

Now let's use the stoichiometry of the reaction to find the moles of N2 produced:

1.735 mol CuO * (1 mol N2 / 3 mol CuO) ≈ 0.5783 mol N2

Now, let's find the mass of N2:

0.5783 mol N2 * (2 * 14) g/mol N2 = 0.5783 mol N2 * 28 g/mol N2 ≈ 16.1924 g N2

Answer: 16.2 g N2

• 4. Let's find the moles of HNO3 consumed:
• 8 g HNO3 * (1 mol HNO3 / 63 g HNO3) ≈ 0.1270 mol HNO3
• Now let's use the stoichiometry of the reaction to find the moles of Mg(NO3)2 produced:
• 0.1270 mol HNO3 * (1 mol Mg(NO3)2 / 2 mol HNO3) ≈ 0.0635 mol Mg(NO3)2
• Now, let's find the mass of Mg(NO3)2:
• 0.0635 mol Mg(NO3)2 * (24 + 2 * (14 + 3 * 16)) g/mol Mg(NO3)2 = 0.0635 mol Mg(NO3)2 * 148 g/mol Mg(NO3)2 ≈ 9.3980 g Mg(NO3)2

Answer: 9.4 g Mg(NO3)2