How many grams of CO2 can be prepared from 150 grams of calcium carbonate (CaCO3) reacting with an excess of hydrochloric (HCl) acid solution?

Explanation

To determine how many grams of CO2 can be prepared from 150 grams of calcium carbonate (CaCO3) reacting with an excess of hydrochloric acid (HCl), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate and hydrochloric acid can be represented as:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the molar mass of calcium carbonate (CaCO3)
The molar mass of CaCO3 is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 3: Convert grams of CaCO3 to moles
To find the number of moles of CaCO3 in 150 grams, use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles of CaCO}_3 = \frac{150 \text{ g}}{100 \text{ g/mol}} = 1.5 \text{ moles} \]

### Step 4: Determine the moles of CO2 produced
From the balanced equation, we see that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 1.5 moles of CaCO3 will produce:
\[ 1.5 \text{ moles of CO}_2 \]

### Step 5: Calculate the mass of CO2 produced
The molar mass of CO2 is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol

Adding these together:
\[ \text{Molar mass of CO}_2 = 12 + 32 = 44 \text{ g/mol} \]

Now, we can calculate the mass of CO2 produced using the number of moles:
\[ \text{Mass of CO}_2 = \text{number of moles} \times \text{molar mass} \]
\[ \text{Mass of CO}_2 = 1.5 \text{ moles} \times 44 \text{ g/mol} = 66 \text{ g} \]

### Final Answer
The mass of CO2 that can be prepared from 150 grams of calcium carbonate is **66 grams**.

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