# 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:**

Molarity of HCl solution = 0.4 M

Volume of solution = 0.5 L

**Putting values in equation 1, we get:**

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

**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 = of calcium carbonate

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

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = 0.1 moles

**Putting values in equation 1, we get:**

**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 =

**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 = container

**Calculating the cost of turns:**

1 container of turns costs $4

So, 0.3125 containers of turns will cost =

**Hence, the cost is coming out to be $ 1.25**

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