How many grams of `MgCO_(3)` contain 24.00 g of oxygen ?
(The molar mass of `MgCO_(3) " is " 84.30 " g mol"^(-1))`

Options

A. 42.15 g
B. 84.30 g
C. 126.00 g
D. 154.00 g

Explanation

To solve the problem of how many grams of \( \text{MgCO}_3 \) contain 24.00 g of oxygen, we can follow these steps:

### Step 1: Determine the mass of oxygen in one mole of \( \text{MgCO}_3 \)
The molar mass of \( \text{MgCO}_3 \) is given as 84.30 g/mol. In one mole of \( \text{MgCO}_3 \), there are three oxygen atoms. The molar mass of oxygen (O) is 16.00 g/mol.

\[ \text{Mass of oxygen in } \text{MgCO}_3 = 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g} \]

### Step 2: Set up a proportion to find the mass of \( \text{MgCO}_3 \) that contains 24.00 g of oxygen
We know that 48.00 g of oxygen corresponds to 84.30 g of \( \text{MgCO}_3 \). We can set up a proportion to find out how much \( \text{MgCO}_3 \) corresponds to 24.00 g of oxygen.

\[ \frac{48.00 \, \text{g O}}{84.30 \, \text{g } \text{MgCO}_3} = \frac{24.00 \, \text{g O}}{x \, \text{g } \text{MgCO}_3} \]

### Step 3: Cross-multiply and solve for \( x \)
Cross-multiplying gives us:

\[ 48.00 \, \text{g O} \cdot x \, \text{g } \text{MgCO}_3 = 24.00 \, \text{g O} \cdot 84.30 \, \text{g } \text{MgCO}_3 \]

This simplifies to:

\[ 48.00x = 24.00 \times 84.30 \]

Calculating the right side:

\[ 24.00 \times 84.30 = 2020.80 \]

So we have:

\[ 48.00x = 2020.80 \]

### Step 4: Solve for \( x \)
Now, divide both sides by 48.00:

\[ x = \frac{2020.80}{48.00} \approx 42.52 \, \text{g} \]

### Conclusion
Thus, the mass of \( \text{MgCO}_3 \) that contains 24.00 g of oxygen is approximately **42.52 g**.

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