The percentage by mass of calcium carbonate in eggshell was determined by adding excess hydrochloric acid to ensure that all the calcium carbonate had reacted. The excess acid left was then titrated with aqueous sodium hydroxide. a) A student added 27.20 cm^3 of 0.200 mol dm^-3 HCl to 0.188 g of eggshell. Calculate the amount, in mol, of HCl added. b) The excess acid requires 23.80 cm^3 of 0.100 dm^-3 NaOH neutralization. Calculate the amount, in mol, of acid that is in excess. c) Determine the amount, in mol, of HCl that reacted with the calcium carbonate in the eggshell. d) State the equation for the reaction of HCl with the calcium carbonate in the eggshell. e) Determined the amount, in mol, of calcium carbonate in the sample of the eggshell. f) Calculate the mass and the percentage by mass of calcium carbonate in the eggshell sample. g) Deduce one assumption made in arriving at the percentage of calcium carbonate in the eggshell sample.
Answers
Answer:
Moles of caco_2= 1/2 moles of Hcl
The mass percentage comes out to be 85.9%
Explanation:
Step 1: To calculate the percentage mass of calcium carbonate in x, shell, we need the data which is given to us, and the data which is given is equal to 1.21 gram. We have 100 moles per liter of solution. The moles of c l are equal to 0.300. The value is equal to 0.03 moles for c n. If you watch the reaction, you'll see that calcium carbonate, solid, plus, 2 c l quas form reacts together to give clo. This is what is going to happen now.
Step 2:When t ml c l is nitrated with calcium carbonate, the moles of the ritter ones will be equal to the moles of the citrate. That gives a point value. It is equal to 0.1 mole volume and a 2 co 3 is equal to 9.23 moles, so this is what we are going to get. The m 1 v 1 formula is equal to 2 m 2. This is also a part of 1 number of moles. That is v 2.
This is for us when we do this. The moles of n, a 2 c o 2 is obese. When we have n na 2 co 2 is equal to basic of n a 23 which is equal to 2 and c. L is going to be acidity of c l which is equal to 1 point from this. It is equal to n n, a 2 c 3, a 23 and a 2 c 3. When we do this, the whole is divided by b c, l, n, h, and c l.
Step 3:The whole will be divided by 20 into 1 point. This is equal to 0.0923 molar. We have found out the molarity for etl. The number of moles for l will be the same as the number of moles for c. The polarity of the universe. We have 0.0923 and the volume was 0.1 liter. We can now calculate the excess remaining moles of excess. The difference will be equal at c l. The value is equal to 0.0277. We will have excess moles at cal. The reaction is going to be calcium carbonate, plus h, c l, s 2, moles of h, c. L. That is very clear. There are 2 mole of c l o right and 1 mole of c 3. A mole of h is equal to a mole of c l and a mole of h is equal to a mole of c l.
Step 4:There are moles of c c 3 in this picture. This is equal to 0.025 when we do this. We are going to get. This is the total number of moles. We're going to get the mass for c c 3. We are aware of the mass. We need to locate this. We need to find the moles so we can calculate the mass. The mass of c a c o 3 is equal to the mass of moles. That is going to be the same as 0.010385 The point is 100.0869. This comes out to be 1.039 gram when we do this.
Step 5:This is the mass of co c 3 in h, shell, so mass percentage, mass percentage of calcium carbonate in axial would be equal to how much is going to be equal to mass of co c 3, divided by the mass of calton 100 point mass of C. The mass of actual is 1.039 and we got it in c 3. We have 1.21, which is 100 point. The mass percentage comes out to be 85.9% when we do this.
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