Respuesta :
Answer: The cost is coming out to be $ 1.25
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of HCl solution = 0.4 M
Volume of solution = 0.5 L
Putting values in equation 1, we get:
[tex]0.4M=\frac{\text{Moles of HCl}}{0.5L}\\\\\text{Moles of HCl}=(0.4mol/L\times 0.5L}=0.2mol[/tex]
The chemical equation for the reaction of HCl and calcium carbonate follows:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
By Stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of calcium carbonate
So, 0.2 moles of HCl will react with = [tex]\frac{1}{2}\times 0.2=0.1mol[/tex] of calcium carbonate
To calculate the mass of calcium carbonate for given moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of calcium carbonate = 100 g/mol
Moles of calcium carbonate = 0.1 moles
Putting values in equation 1, we get:
[tex]0.1mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(0.1mol\times 100g/mol)=10g[/tex]
- Calculating the mass of calcium carbonate in 1 container:
We are given:
One container contains eighty 1 g of tablets, this means that in total 80 g of tablets are there.
Every container has 40 % calcium carbonate.
Mass of calcium carbonate in 1 container = 40 % of 80 g = [tex]\frac{40}{100}\times 80=32g[/tex]
- Calculating the containers for amount of calcium carbonate that neutralized HCl by using unitary method:
32 grams of calcium carbonate is present in 1 container
So, 10 g of calcium carbonate will be present in = [tex]\frac{1}{32}\times 10=0.3125[/tex] container
- Calculating the cost of turns:
1 container of turns costs $4
So, 0.3125 containers of turns will cost = [tex]\frac{\$ 4}{1}\times 0.3125=\$ 1.25[/tex]
Hence, the cost is coming out to be $ 1.25
