2. 25.0mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg2 requires 2.12mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2 is added requires 25.88mL of the EDTA to reach the end point. a. How many mL of EDTA are needed to titrate the Ca2 ion in the aliquot

2) 25.0mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg2+ requires 2.12mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 25.88mL of the EDTA to reach the end point.

a. How many mL of EDTA are needed to titrate the Ca2+ ion in the aliquot?

b. How many moles of EDTA are there in the volume obtained in a.?

c. What is the molarity of the EDTA solution?

Explanation:

Given that;

Volume of aliquot = 25mL

Blank reading = 2.12mL

2a)

Volume of EDTA used for Ca²⁺ ion

25.88mL - 2.12mL = 23.76mL

Therefore mL of EDTA needed to titrate the Ca²⁺ ion in the aliquot is 23.76mL

2b)

Molarity of Ca²⁺ ion  is 0.0172M

Mole of EDTA =

2c)

Molarity of EDTA = mole of EDTA / Vol. of EDTA

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

c) should be the answer.