Chemistry worksheet focusing on stoichiometry with multiple problems requiring balancing equations and performing mole-to-mass conversions.
Chemistry worksheet on stoichiometry featuring problems involving balancing chemical equations and calculating moles, grams, and masses for various reactions.
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Mass to Mole Stoichiometry Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Mass to Mole Stoichiometry Worksheets
Here are the step-by-step solutions for the chemistry worksheet.
First, we must balance the chemical equation:
$$4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$$
a. How many moles of NO are formed if 824 g of $\text{NH}_3$ react?
* Step 1: Find the molar mass of ammonia ($\text{NH}_3$). Nitrogen is 14 and Hydrogen is 1. So, $14 + (3 \times 1) = 17 \text{ g/mol}$.
* Step 2: Convert grams to moles. $824 \text{ g} / 17 \text{ g/mol} \approx 48.47 \text{ moles}$ of $\text{NH}_3$.
* Step 3: Look at the balanced equation. The ratio of $\text{NH}_3$ to $\text{NO}$ is 4 to 4 (which is a 1:1 ratio).
* Step 4: Therefore, 48.47 moles of $\text{NH}_3$ will produce 48.5 moles of $\text{NO}$.
b. How many grams of water are formed if 2.55 mol of ammonia are oxidized?
* Step 1: Look at the mole ratio in the balanced equation. 4 moles of $\text{NH}_3$ produce 6 moles of $\text{H}_2\text{O}$. This simplifies to a ratio of 1.5 (since $6/4 = 1.5$).
* Step 2: Calculate moles of water. $2.55 \text{ mol NH}_3 \times 1.5 = 3.825 \text{ moles H}_2\text{O}$.
* Step 3: Find the molar mass of water ($\text{H}_2\text{O}$). $(2 \times 1) + 16 = 18 \text{ g/mol}$.
* Step 4: Convert moles to grams. $3.825 \text{ mol} \times 18 \text{ g/mol} = 68.85 \text{ g}$. Rounding to 3 significant figures gives 68.9 g.
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Equation: $2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2$
a. How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
* Step 1: Find the molar mass of Oxygen gas ($\text{O}_2$). $16 \times 2 = 32 \text{ g/mol}$.
* Step 2: Convert grams of oxygen to moles. $125 \text{ g} / 32 \text{ g/mol} = 3.906 \text{ moles O}_2$.
* Step 3: Use the mole ratio. The equation shows that 2 moles of $\text{HgO}$ are needed for every 1 mole of $\text{O}_2$.
* Step 4: Multiply by 2. $3.906 \times 2 = 7.812$. Rounding to 3 significant figures gives 7.81 moles.
b. How many grams of mercury are produced if 2.45 moles of mercury (II) oxide decompose?
* Step 1: Check the mole ratio. The equation shows a 2:2 ratio between $\text{HgO}$ and $\text{Hg}$. This means they are equal. So, 2.45 moles of $\text{HgO}$ produce 2.45 moles of $\text{Hg}$.
* Step 2: Find the atomic mass of Mercury ($\text{Hg}$). It is approximately $200.59 \text{ g/mol}$.
* Step 3: Convert moles to grams. $2.45 \text{ mol} \times 200.59 \text{ g/mol} = 491.44 \text{ g}$. Rounding to 3 significant figures gives 491 g.
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Equation: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$
a. How many grams of HCl are consumed by the reaction of 2.5 moles of magnesium?
* Step 1: Check the mole ratio. For every 1 mole of $\text{Mg}$, you need 2 moles of $\text{HCl}$.
* Step 2: Calculate moles of $\text{HCl}$. $2.5 \text{ mol Mg} \times 2 = 5.0 \text{ moles HCl}$.
* Step 3: Find the molar mass of $\text{HCl}$. $1 (\text{H}) + 35.5 (\text{Cl}) = 36.5 \text{ g/mol}$.
* Step 4: Convert moles to grams. $5.0 \text{ mol} \times 36.5 \text{ g/mol} =$ 182.5 g.
b. What is the mass in grams of $\text{H}_2$ gas when 4 moles of HCl is added to the reaction?
* Step 1: Check the mole ratio. 2 moles of $\text{HCl}$ produce 1 mole of $\text{H}_2$.
* Step 2: Calculate moles of $\text{H}_2$. $4 \text{ mol HCl} / 2 = 2 \text{ moles H}_2$.
* Step 3: Find the molar mass of Hydrogen gas ($\text{H}_2$). $1 \times 2 = 2 \text{ g/mol}$.
* Step 4: Convert moles to grams. $2 \text{ mol} \times 2 \text{ g/mol} =$ 4 g.
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Equation: $\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
a. If 3.2 moles of $\text{CaC}_2$ are consumed, how many grams of $\text{H}_2\text{O}$ are needed?
* Step 1: Check the mole ratio. 1 mole of $\text{CaC}_2$ needs 2 moles of $\text{H}_2\text{O}$.
* Step 2: Calculate moles of water needed. $3.2 \text{ mol} \times 2 = 6.4 \text{ moles H}_2\text{O}$.
* Step 3: Find the molar mass of water ($\text{H}_2\text{O}$). $18 \text{ g/mol}$.
* Step 4: Convert moles to grams. $6.4 \text{ mol} \times 18 \text{ g/mol} =$ 115.2 g.
b. How many grams of $\text{Ca(OH)}_2$ would be formed with 3.2 moles of $\text{CaC}_2$?
* Step 1: Check the mole ratio. 1 mole of $\text{CaC}_2$ produces 1 mole of $\text{Ca(OH)}_2$. So, we get 3.2 moles of product.
* Step 2: Find the molar mass of Calcium Hydroxide ($\text{Ca(OH)}_2$).
* Ca: 40
* O: $16 \times 2 = 32$
* H: $1 \times 2 = 2$
* Total: $40 + 32 + 2 = 74 \text{ g/mol}$.
* Step 3: Convert moles to grams. $3.2 \text{ mol} \times 74 \text{ g/mol} =$ 236.8 g.
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Final Answer:
1.
a. 48.5 moles of NO
b. 68.9 g of water
2.
a. 7.81 moles of HgO
b. 491 g of Mercury
3.
a. 182.5 g of HCl
b. 4 g of H₂
4.
a. 115.2 g of H₂O
b. 236.8 g of Ca(OH)₂
1. Ammonia Reaction
First, we must balance the chemical equation:
$$4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$$
a. How many moles of NO are formed if 824 g of $\text{NH}_3$ react?
* Step 1: Find the molar mass of ammonia ($\text{NH}_3$). Nitrogen is 14 and Hydrogen is 1. So, $14 + (3 \times 1) = 17 \text{ g/mol}$.
* Step 2: Convert grams to moles. $824 \text{ g} / 17 \text{ g/mol} \approx 48.47 \text{ moles}$ of $\text{NH}_3$.
* Step 3: Look at the balanced equation. The ratio of $\text{NH}_3$ to $\text{NO}$ is 4 to 4 (which is a 1:1 ratio).
* Step 4: Therefore, 48.47 moles of $\text{NH}_3$ will produce 48.5 moles of $\text{NO}$.
b. How many grams of water are formed if 2.55 mol of ammonia are oxidized?
* Step 1: Look at the mole ratio in the balanced equation. 4 moles of $\text{NH}_3$ produce 6 moles of $\text{H}_2\text{O}$. This simplifies to a ratio of 1.5 (since $6/4 = 1.5$).
* Step 2: Calculate moles of water. $2.55 \text{ mol NH}_3 \times 1.5 = 3.825 \text{ moles H}_2\text{O}$.
* Step 3: Find the molar mass of water ($\text{H}_2\text{O}$). $(2 \times 1) + 16 = 18 \text{ g/mol}$.
* Step 4: Convert moles to grams. $3.825 \text{ mol} \times 18 \text{ g/mol} = 68.85 \text{ g}$. Rounding to 3 significant figures gives 68.9 g.
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2. Mercury (II) Oxide Decomposition
Equation: $2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2$
a. How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
* Step 1: Find the molar mass of Oxygen gas ($\text{O}_2$). $16 \times 2 = 32 \text{ g/mol}$.
* Step 2: Convert grams of oxygen to moles. $125 \text{ g} / 32 \text{ g/mol} = 3.906 \text{ moles O}_2$.
* Step 3: Use the mole ratio. The equation shows that 2 moles of $\text{HgO}$ are needed for every 1 mole of $\text{O}_2$.
* Step 4: Multiply by 2. $3.906 \times 2 = 7.812$. Rounding to 3 significant figures gives 7.81 moles.
b. How many grams of mercury are produced if 2.45 moles of mercury (II) oxide decompose?
* Step 1: Check the mole ratio. The equation shows a 2:2 ratio between $\text{HgO}$ and $\text{Hg}$. This means they are equal. So, 2.45 moles of $\text{HgO}$ produce 2.45 moles of $\text{Hg}$.
* Step 2: Find the atomic mass of Mercury ($\text{Hg}$). It is approximately $200.59 \text{ g/mol}$.
* Step 3: Convert moles to grams. $2.45 \text{ mol} \times 200.59 \text{ g/mol} = 491.44 \text{ g}$. Rounding to 3 significant figures gives 491 g.
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3. Magnesium and HCl Reaction
Equation: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$
a. How many grams of HCl are consumed by the reaction of 2.5 moles of magnesium?
* Step 1: Check the mole ratio. For every 1 mole of $\text{Mg}$, you need 2 moles of $\text{HCl}$.
* Step 2: Calculate moles of $\text{HCl}$. $2.5 \text{ mol Mg} \times 2 = 5.0 \text{ moles HCl}$.
* Step 3: Find the molar mass of $\text{HCl}$. $1 (\text{H}) + 35.5 (\text{Cl}) = 36.5 \text{ g/mol}$.
* Step 4: Convert moles to grams. $5.0 \text{ mol} \times 36.5 \text{ g/mol} =$ 182.5 g.
b. What is the mass in grams of $\text{H}_2$ gas when 4 moles of HCl is added to the reaction?
* Step 1: Check the mole ratio. 2 moles of $\text{HCl}$ produce 1 mole of $\text{H}_2$.
* Step 2: Calculate moles of $\text{H}_2$. $4 \text{ mol HCl} / 2 = 2 \text{ moles H}_2$.
* Step 3: Find the molar mass of Hydrogen gas ($\text{H}_2$). $1 \times 2 = 2 \text{ g/mol}$.
* Step 4: Convert moles to grams. $2 \text{ mol} \times 2 \text{ g/mol} =$ 4 g.
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4. Acetylene Production
Equation: $\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
a. If 3.2 moles of $\text{CaC}_2$ are consumed, how many grams of $\text{H}_2\text{O}$ are needed?
* Step 1: Check the mole ratio. 1 mole of $\text{CaC}_2$ needs 2 moles of $\text{H}_2\text{O}$.
* Step 2: Calculate moles of water needed. $3.2 \text{ mol} \times 2 = 6.4 \text{ moles H}_2\text{O}$.
* Step 3: Find the molar mass of water ($\text{H}_2\text{O}$). $18 \text{ g/mol}$.
* Step 4: Convert moles to grams. $6.4 \text{ mol} \times 18 \text{ g/mol} =$ 115.2 g.
b. How many grams of $\text{Ca(OH)}_2$ would be formed with 3.2 moles of $\text{CaC}_2$?
* Step 1: Check the mole ratio. 1 mole of $\text{CaC}_2$ produces 1 mole of $\text{Ca(OH)}_2$. So, we get 3.2 moles of product.
* Step 2: Find the molar mass of Calcium Hydroxide ($\text{Ca(OH)}_2$).
* Ca: 40
* O: $16 \times 2 = 32$
* H: $1 \times 2 = 2$
* Total: $40 + 32 + 2 = 74 \text{ g/mol}$.
* Step 3: Convert moles to grams. $3.2 \text{ mol} \times 74 \text{ g/mol} =$ 236.8 g.
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Final Answer:
1.
a. 48.5 moles of NO
b. 68.9 g of water
2.
a. 7.81 moles of HgO
b. 491 g of Mercury
3.
a. 182.5 g of HCl
b. 4 g of H₂
4.
a. 115.2 g of H₂O
b. 236.8 g of Ca(OH)₂
Parent Tip: Review the logic above to help your child master the concept of chemistry stoichiometry worksheet.
