1.250 g of metal carbonate `(MCO_(3))` was treated with 500 mL of 0.1 M HCl solution. The unreacted HCl required 50.0 mL of 0.500 M NaOH solution for neutralization, Identify the metal M

Options

A. Mg
B. Ca
C. Sr
D. Ba

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the initial moles of HCl
We start with 500 mL of 0.1 M HCl. To find the number of moles of HCl, we use the formula:

\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]

Converting 500 mL to liters:

\[ 500 \, \text{mL} = 0.500 \, \text{L} \]

Now, calculating the moles of HCl:

\[ \text{Moles of HCl} = 0.500 \, \text{L} \times 0.1 \, \text{mol/L} = 0.050 \, \text{mol} = 50 \, \text{mmol} \]

### Step 2: Calculate the moles of NaOH used
Next, we have 50 mL of 0.500 M NaOH used for neutralization. We calculate the moles of NaOH:

\[ \text{Volume (L)} = 50 \, \text{mL} = 0.050 \, \text{L} \]

\[ \text{Moles of NaOH} = 0.050 \, \text{L} \times 0.500 \, \text{mol/L} = 0.025 \, \text{mol} = 25 \, \text{mmol} \]

### Step 3: Determine the moles of HCl that reacted with MCO3
The total moles of HCl initially were 50 mmol, and 25 mmol were neutralized by NaOH. Therefore, the moles of HCl that reacted with the metal carbonate (MCO3) are:

\[ \text{Moles of HCl reacted} = 50 \, \text{mmol} - 25 \, \text{mmol} = 25 \, \text{mmol} \]

### Step 4: Relate moles of HCl to moles of MCO3
From the reaction of MCO3 with HCl:

\[ \text{MCO}_3 + 2 \text{HCl} \rightarrow \text{MCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]

1 mole of MCO3 reacts with 2 moles of HCl. Therefore, the moles of MCO3 that reacted can be calculated as:

\[ \text{Moles of MCO3} = \frac{25 \, \text{mmol}}{2} = 12.5 \, \text{mmol} \]

### Step 5: Calculate the molar mass of MCO3
We know the mass of MCO3 is 1.250 g. Using the moles calculated, we can find the molar mass:

\[ \text{Molar mass of MCO3} = \frac{\text{mass}}{\text{moles}} = \frac{1.250 \, \text{g}}{12.5 \times 10^{-3} \, \text{mol}} = 100 \, \text{g/mol} \]

### Step 6: Identify the metal M
The molar mass of MCO3 is 100 g/mol. The molar mass can be expressed as:

\[ \text{Molar mass of MCO3} = \text{M} + 12 + 3 \times 16 \]

Calculating the mass of CO3:

\[ \text{Molar mass of CO3} = 12 + 48 = 60 \, \text{g/mol} \]

Setting up the equation:

\[ \text{M} + 60 = 100 \]

Solving for M:

\[ \text{M} = 100 - 60 = 40 \, \text{g/mol} \]

### Conclusion
The metal M has a molar mass of 40 g/mol, which corresponds to calcium (Ca). Therefore, the metal M is **calcium**.

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