The reduction of 1.49 of a metal oxide required 560 ml of `H_(2)` at STP. If atomic mass of metal is 40, formula of its chloride will be `MCl_(x)x=`______________.

Answer

To solve the problem step by step, we need to determine the formula of the chloride of the metal (MClₓ) based on the information provided. Here’s how we can approach it:

### Step 1: Calculate the number of moles of H₂ used
Given that 560 ml of H₂ is used at STP (Standard Temperature and Pressure), we can convert this volume to moles.

At STP, 1 mole of gas occupies 22.4 liters (or 22400 ml).

\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2 \text{ (in ml)}}{22400 \text{ ml/mol}} = \frac{560 \text{ ml}}{22400 \text{ ml/mol}} = 0.025 \text{ moles} \]

### Step 2: Determine the number of moles of electrons transferred
Since each mole of H₂ provides 2 moles of electrons (as H₂ → 2H⁺ + 2e⁻), the total number of moles of electrons transferred is:

\[ \text{Moles of electrons} = 0.025 \text{ moles of } H_2 \times 2 = 0.05 \text{ moles of electrons} \]

### Step 3: Calculate the equivalent weight of the metal oxide
The reduction of 1.49 g of the metal oxide corresponds to the loss of 0.05 moles of electrons. The equivalent weight (E) can be calculated using the formula:

\[ E = \frac{\text{mass of metal oxide}}{\text{moles of electrons}} = \frac{1.49 \text{ g}}{0.05} = 29.8 \text{ g/equiv} \]

### Step 4: Determine the molar mass of the metal
The equivalent weight is related to the molar mass (M) of the metal by the formula:

\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]

Where \( n \) is the number of electrons lost per atom of the metal during the reduction. We can find \( n \) using the equivalent weight calculated:

\[ \text{Molar mass} = \text{Equivalent weight} + \text{Molar mass of oxygen (16 g/mol)} = 29.8 + 16 = 45.8 \text{ g/mol} \]

Given that the atomic mass of the metal is 40 g/mol, we can find \( n \):

\[ n = \frac{\text{Molar mass}}{\text{Equivalent weight}} = \frac{40}{12.8} \approx 3.125 \approx 3 \text{ (rounding to the nearest whole number)} \]

### Step 5: Write the formula for the chloride
Since \( n \) is approximately 3, the formula for the chloride of the metal will be:

\[ \text{MCl}_3 \]

### Final Answer
The formula of the chloride is \( MCl_3 \).

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