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To determine the mass of 90% pure calcium carbonate (CaCO₃) required to neutralize 2 liters of decinormal HCl solution, we need to follow a series of steps based on stoichiometry and given data.
### Step-by-Step Solution:
1. Identify the reaction:
The chemical equation for the neutralization reaction is:
[tex]\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \][/tex]
This means that one mole of CaCO₃ reacts with two moles of HCl.
2. Calculate moles of HCl:
We know the volume and molarity of HCl. Molarity (M) is defined as moles of solute per liter of solution.
[tex]\[ \text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl in liters} \][/tex]
Given:
[tex]\[ \text{Molarity of HCl} = 0.1 \, \text{M} \][/tex]
[tex]\[ \text{Volume of HCl} = 2 \, \text{liters} \][/tex]
[tex]\[ \text{Moles of HCl} = 0.1 \, \text{M} \times 2 \, \text{liters} = 0.2 \, \text{moles} \][/tex]
3. Determine moles of CaCO₃ needed:
According to the stoichiometry of the balanced reaction, 1 mole of CaCO₃ is required for every 2 moles of HCl.
[tex]\[ \text{Moles of CaCO}_3 \, \text{needed} = \frac{\text{Moles of HCl}}{2} \][/tex]
[tex]\[ \text{Moles of CaCO}_3 \, \text{needed} = \frac{0.2 \, \text{moles}}{2} = 0.1 \, \text{moles} \][/tex]
4. Calculate the mass of pure CaCO₃:
The molar mass of CaCO₃ is approximately 100.09 g/mol.
[tex]\[ \text{Mass of pure CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]
[tex]\[ \text{Mass of pure CaCO}_3 = 0.1 \, \text{moles} \times 100.09 \, \text{g/mol} = 10.009 \, \text{g} \][/tex]
5. Adjust for the purity of CaCO₃:
Since the CaCO₃ is 90% pure, we need more than the calculated pure mass to provide the required amount of pure CaCO₃.
[tex]\[ \text{Mass of 90% pure CaCO}_3 = \frac{\text{Mass of pure CaCO}_3}{\text{Purity fraction}} \][/tex]
Given the purity as 90%, which is 0.90 in decimal form:
[tex]\[ \text{Mass of 90% pure CaCO}_3 = \frac{10.009 \, \text{g}}{0.90} \approx 11.121 \, \text{g} \][/tex]
To summarize, the mass of 90% pure CaCO₃ required to neutralize 2 liters of decinormal HCl solution is approximately 11.121 g.
### Step-by-Step Solution:
1. Identify the reaction:
The chemical equation for the neutralization reaction is:
[tex]\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \][/tex]
This means that one mole of CaCO₃ reacts with two moles of HCl.
2. Calculate moles of HCl:
We know the volume and molarity of HCl. Molarity (M) is defined as moles of solute per liter of solution.
[tex]\[ \text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl in liters} \][/tex]
Given:
[tex]\[ \text{Molarity of HCl} = 0.1 \, \text{M} \][/tex]
[tex]\[ \text{Volume of HCl} = 2 \, \text{liters} \][/tex]
[tex]\[ \text{Moles of HCl} = 0.1 \, \text{M} \times 2 \, \text{liters} = 0.2 \, \text{moles} \][/tex]
3. Determine moles of CaCO₃ needed:
According to the stoichiometry of the balanced reaction, 1 mole of CaCO₃ is required for every 2 moles of HCl.
[tex]\[ \text{Moles of CaCO}_3 \, \text{needed} = \frac{\text{Moles of HCl}}{2} \][/tex]
[tex]\[ \text{Moles of CaCO}_3 \, \text{needed} = \frac{0.2 \, \text{moles}}{2} = 0.1 \, \text{moles} \][/tex]
4. Calculate the mass of pure CaCO₃:
The molar mass of CaCO₃ is approximately 100.09 g/mol.
[tex]\[ \text{Mass of pure CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]
[tex]\[ \text{Mass of pure CaCO}_3 = 0.1 \, \text{moles} \times 100.09 \, \text{g/mol} = 10.009 \, \text{g} \][/tex]
5. Adjust for the purity of CaCO₃:
Since the CaCO₃ is 90% pure, we need more than the calculated pure mass to provide the required amount of pure CaCO₃.
[tex]\[ \text{Mass of 90% pure CaCO}_3 = \frac{\text{Mass of pure CaCO}_3}{\text{Purity fraction}} \][/tex]
Given the purity as 90%, which is 0.90 in decimal form:
[tex]\[ \text{Mass of 90% pure CaCO}_3 = \frac{10.009 \, \text{g}}{0.90} \approx 11.121 \, \text{g} \][/tex]
To summarize, the mass of 90% pure CaCO₃ required to neutralize 2 liters of decinormal HCl solution is approximately 11.121 g.
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