During a lab activity, students prepared two solutions separately at the same temperature. Later they mixed the solutions and th
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<u>Answer:</u> The mass percent of calcium in milk is 0.107 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of calcium oxalate = 0.429 g
Molar mass of calcium oxalate = 128.1 g/mol
Putting values in equation 1, we get:
The given chemical equation follows:
Sodium oxalate is present in excess. So, it is considered as an excess reagent. And, calcium ion is a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of calcium oxalate is produced from 1 mole of calcium ion
So, 0.0033 moles of calcium oxalate is produced from = of calcium ions
- Now, calculating the mass of calcium ions by using equation 1, we get:
Moles of calcium ions = 0.0033 moles
Molar mass of calcium ions = 40 g/mol
Putting values in equation 1, we get:
- To calculate the mass percentage of calcium ions in milk, we use the equation:
Mass of milk = 125 g
Mass of calcium ions = 0.132 g
Putting values in above equation, we get:
Hence, the mass percent of calcium in milk is 0.107 %
<u>Answer:</u> The molarity of barium hydroxide solution is 0.118 M.
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
Hence, the molarity of solution will be 0.118 M.