Question

Given the unbalanced equation (Reaction 1) and the molecular weight of calcium citrate (498.5 g/mol), if 15 nmol of calcium oxalate is mixed with 15 nmol of potassium citrate, what is the approximate theoretical yield of calcium citrate? 3 CaC₂O₄ + 2 K₃C₆H₅O₇ → Ca₃(C₆H₅O₇)₂ + 3K₂C₂O₄ A. 1,250 ng B. 2,500 ng C. 3,750 ng D. 7,500 ng

Answer 1

The theoretical yield of calcium citrate obtained when 15 nmole of calcium oxalate is mixed with 15 nmole of potassium citrate is 2500 ng (option B)

How to calculate the theoretical yield of calcium citrate?

The theoretical yield of calcium citrate can be calculated as follow:

1. Determine the limiting reactant.

Balanced equation

`3CaC_2O_4\ +\ 2K_3C_6H_5O_7\ \rightarrow Ca_3(C_6H_5O_7)_2\ +\ 3K_2C_2O_4`

From the equation,

3 nmoles of
`CaC_2O_4` reacted with 2 nmoles of
`K_3C_6H_5O_7`

Therefore,

15 nmoles of
`CaC_2O_4` will react with =
`(15\ *\ 2)/(3)` = 10 nmoles of
`K_3C_6H_5O_7`

From the above, all the amount of
`CaC_2O_4` took part in the reaction while only 10 nmole of
`K_3C_6H_5O_7` reacted.

Thus,
`CaC_2O_4` is the limiting reactant

2. Determine the nmole of calcium citrate obtained.

`3CaC_2O_4\ +\ 2K_3C_6H_5O_7\ \rightarrow Ca_3(C_6H_5O_7)_2\ +\ 3K_2C_2O_4`

From the equation,

3 nmoles of
`CaC_2O_4` reacted to produce 1 nmole of
`Ca_3(C_6H_5O_7)_2`

Therefore,

15 nmoles of
`CaC_2O_4` will react to produce =
`(15)/(3)` = 5 nmoles of
`Ca_3(C_6H_5O_7)_2`

3. Now, we can calculate the theoretical yield of calcium citrate. Details below:

• Molar mass of calcium citrate = 498.5 g/mol
• Mole of of calcium citrate = 5 nmoles
• Theoretical yield of calcium citrate =?

Theoretical yield of calcium citrate = Mole × molar mass

= 5 × 498.5

≈ 2500 ng

Thus, the answer to the question is option B