One container of turns costs 4 dollars. Each container has eighty 1g tablets. Assume each turns is 40% caco₃. Using only turns, you are required to neutrailize 0.5L of 0.4M hcl. How much does this cost?

Answer: The cost is coming out to be $ 1.25

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl solution = 0.4 M

Volume of solution = 0.5 L

Putting values in equation 1, we get:

$0.4M=\frac{\text{Moles of HCl}}{0.5L}\\\\\text{Moles of HCl}=(0.4mol/L\times 0.5L}=0.2mol$

The chemical equation for the reaction of HCl and calcium carbonate follows:

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

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 = $\frac{1}{2}\times 0.2=0.1mol$ of calcium carbonate

To calculate the mass of calcium carbonate for given moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = 0.1 moles

Putting values in equation 1, we get:

$0.1mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(0.1mol\times 100g/mol)=10g$

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 = $\frac{40}{100}\times 80=32g$

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 = $\frac{1}{32}\times 10=0.3125$ container

1 container of turns costs $4

So, 0.3125 containers of turns will cost = $\frac{\$ 4}{1}\times 0.3125=\$ 1.25$

Hence, the cost is coming out to be $ 1.25

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