# 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