To solve the problem step-by-step, we will follow the chemical reactions and calculations as described in the video transcript.
### Step 1: Write the reaction
When `M2CO3` reacts with hydrochloric acid (HCl), the reaction can be represented as:
\[ \text{M}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{MCl} + \text{H}_2\text{O} + \text{CO}_2 \]
### Step 2: Calculate the moles of HCl used
The total volume of HCl used is 150 mL with a normality of 1 N. First, we convert the volume from mL to L:
\[ 150 \, \text{mL} = 0.150 \, \text{L} \]
Now, we can calculate the moles of HCl:
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume in L} = 1 \, \text{N} \times 0.150 \, \text{L} = 0.150 \, \text{moles} \]
### Step 3: Calculate the moles of NaOH used
The unused HCl required 100 mL of 0.5 N NaOH. First, we convert the volume from mL to L:
\[ 100 \, \text{mL} = 0.100 \, \text{L} \]
Now, we can calculate the moles of NaOH:
\[ \text{Moles of NaOH} = \text{Normality} \times \text{Volume in L} = 0.5 \, \text{N} \times 0.100 \, \text{L} = 0.050 \, \text{moles} \]
### Step 4: Calculate the moles of HCl that reacted with NaOH
Since the stoichiometry of the reaction between HCl and NaOH is 1:1, the moles of HCl that reacted with NaOH is also 0.050 moles.
### Step 5: Calculate the moles of HCl that reacted with M2CO3
The total moles of HCl was 0.150 moles, and the moles that reacted with NaOH was 0.050 moles. Therefore, the moles of HCl that reacted with `M2CO3` is:
\[ \text{Moles of HCl reacting with M2CO3} = 0.150 - 0.050 = 0.100 \, \text{moles} \]
### Step 6: Relate moles of HCl to moles of M2CO3
From the balanced equation, 2 moles of HCl react with 1 mole of `M2CO3`. Thus, we can find the moles of `M2CO3` that reacted:
\[ \text{Moles of M2CO3} = \frac{0.100}{2} = 0.050 \, \text{moles} \]
### Step 7: Calculate the molar mass of M2CO3
We know the mass of `M2CO3` is 5.3 g. Using the formula for moles:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
We can rearrange this to find the molar mass:
\[ \text{Molar mass of M2CO3} = \frac{5.3 \, \text{g}}{0.050 \, \text{moles}} = 106 \, \text{g/mol} \]
### Step 8: Set up the equation for molar mass
The molar mass of `M2CO3` can be expressed as:
\[ 2M + 12 + 48 = 106 \]
Where 12 is the atomic mass of carbon and 48 is the total mass of three oxygen atoms (3 × 16).
### Step 9: Solve for M
Rearranging the equation:
\[ 2M + 60 = 106 \]
\[ 2M = 46 \]
\[ M = 23 \, \text{g/mol} \]
### Step 10: Calculate the equivalent mass of M
The equivalent mass is calculated as:
\[ \text{Equivalent mass} = \frac{\text{Atomic mass}}{\text{Valency}} \]
Since the valency of M in `M2CO3` is 1 (as derived from the charge balance), we have:
\[ \text{Equivalent mass} = \frac{23}{1} = 23 \, \text{g/equiv} \]
### Final Answer
The equivalent mass of M is **23 g/equiv**.
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