Unit 8 Stoichiometry Review

2H₂  +   O₂  →  2H₂O
How many moles of oxygen are consumed if 8 moles H₂ are used?

• 2

• 4

• 6

• 8

Answer explanation

According to the balanced equation, 2 moles of H2 react with 1 mole of O2. If 8 moles of H2 are used, the ratio is 8:2, which means 4 moles of O2 are consumed. Thus, the correct answer is 4.

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NGSS.HS-PS1-7

2 KClO₃ → 2 KCl + 3 O₂ How many moles of oxygen are produced when 6.7 moles of KClO₃ react?

• 6.7 mol

• 1.0 mol

• 10.1 mol

• 4.5 mol

Answer explanation

From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, 6.7 moles of KClO3 will produce (6.7 moles KClO3) * (3 moles O2 / 2 moles KClO3) = 10.1 moles O2.

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NGSS.HS-PS1-7

2Na + 2H₂O → 2NaOH+ H₂ How many grams of hydrogen are produced if 120 g of Na are available? The molar mass of Na is 23 g. The molar mass of H2 is 2 g.

• 5.2 g

• 2.6 g

• 690 g

• 45 g

Answer explanation

First, calculate moles of Na: 120 g / 23 g/mol = 5.22 mol. The reaction shows 2 moles of Na produce 1 mole of H2. Thus, 5.22 mol Na produces 5.22/2 = 2.61 mol H2. Convert to grams: 2.61 mol * 2 g/mol = 5.2 g H2.

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NGSS.HS-PS1-7

What is stoichiometry?

• Whether a precipitate is formed in a chemical reaction

• Whether a reaction is endothermic or exothermic

• How quantities of reactants and quantities of products are related

• How many chemical changes each reaction goes through

Answer explanation

Stoichiometry is the study of how quantities of reactants and products are related in a chemical reaction, allowing us to predict the amounts of substances consumed and produced.

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NGSS.HS-PS1-7

CH₄ + 2 O₂ → CO₂ + 2 H₂O How many moles of carbon dioxide are produced from the combustion of 110 g of CH₄? The molar mass of CH4 is 16 g.

• 13.7 mol

• 2.75 mol

• 6.11 mol

• 6.88 mol

Answer explanation

To find moles of CO2 produced, first calculate moles of CH4: 110 g / 16 g/mol = 6.875 mol CH4. The reaction shows 1 mol CH4 produces 1 mol CO2, so 6.875 mol CH4 produces 6.875 mol CO2, rounded to 6.88 mol.

Which is true due to the Law of Conservation of Matter?

• Mass of reactants < mass of products

• Mass of reactants > mass of products

• Mass of reactants = mass of products

• Mass of reactants and mass of products are unrelated.

Answer explanation

The Law of Conservation of Matter states that matter cannot be created or destroyed in a chemical reaction. Therefore, the mass of reactants must equal the mass of products, making 'Mass of reactants = mass of products' the correct choice.

Tags

NGSS.HS-PS1-7

4 Al + 3 O₂ –> 2 Al₂O₃ 

How many moles of aluminum would be needed to react with 45 grams of O₂? The molar mass of O2 is 32 g.

• 1.05 moles

• 3.75 moles

• 2.06 moles

• 1.875 moles

Answer explanation

To find moles of Al needed, first calculate moles of O2: 45 g / 32 g/mol = 1.40625 moles. From the reaction, 3 moles of O2 react with 4 moles of Al. Thus, (4/3) * 1.40625 = 1.875 moles of Al needed.

Tags

NGSS.HS-PS1-7

Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l)

How many moles of Fe can be made from 135 L H₂? 

• 4.02 moles Fe

• 6.99 moles Fe

• 9.13 moles Fe

• 2.81 moles Fe

Answer explanation

From the reaction, 3 moles of H2 produce 2 moles of Fe. First, convert 135 L of H2 to moles using the ideal gas law (22.4 L/mol at STP): 135 L / 22.4 L/mol = 6.02 moles H2. Then, (6.02 moles H2) * (2 moles Fe / 3 moles H2) = 4.02 moles Fe.

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂

How many grams of Al would react with 2.55 mol H₂SO₄? The molar mass of Al is 27 g.

• 15.6 g

• 45.9 g

• 459 g

• 870 g

Answer explanation

From the balanced equation, 2 moles of Al react with 3 moles of H2SO4. For 2.55 mol H2SO4, (2.55 mol H2SO4) * (2 mol Al / 3 mol H2SO4) = 1.70 mol Al. Thus, 1.70 mol Al * 27 g/mol = 45.9 g Al.

Use the equation 2 Al + 3 Cl₂ ---> 2 AlCl₃.  If 2 moles of aluminum and 2 moles of chlorine are reacted, identify the limiting reactant.

• AlCl3

• Cl2

• Al

Answer explanation

In the reaction, 2 moles of Al require 3 moles of Cl2. With only 2 moles of Cl2 available, it is insufficient to react with 2 moles of Al. Therefore, Cl2 is the limiting reactant.

Tags

NGSS.HS-PS1-7