Write the balanced equation for the reaction of potassium carbonate and hydrochl

Question

Write the balanced equation for the reaction of potassium carbonate and hydrochloric acid. What is the excess reagent in this experiment? What happens to that excess reagent at the end of the experiment? Calculate the mass of LiCi that would be produced if 1,50000 grams of Li_3CO_3 reacts with excess HC^1. Calculate the mass of MgCO_3 that would be required to produce 1,0000 grams of MgCl_2. If only 4,000M HCI was available for you to use, calculate the volume of 4,000M solution required to completely react with 1,700 grams of Na_2CO_3.

Explanation / Answer

1) Molecular formula of Potassium carbonate is K2CO3 and of Hydrochloric acid is HCl.

Net balance reaction between K2CO3 and HCl is

K2CO3 + 2 HCl -----------> 2 KCl + H2CO3 .

There is formation of Carbonic acid (H2CO3) and potassium chloride.

====================================================

2) Practical detail required.

===================================================

3) Balance equation for the formation of LiCl from Li2CO3 and HCl is,

Li2CO3 + 2HCl ----------> 2LiCl + H2CO3.

Stochiometry says,

1 mole of Li2CO3 gives 2 moles of LiCl.

I.e. equivalence is given as,

1 mole of Li2CO3 2 moles of LiCl.

Molar mass of Li2CO3 = 73.89 g/mol

Molar mass of LiCl = 42.39 g/mol.

Hence mass equivalence is written as,

73.89 g of Li2CO3 2 x 42.39 g of LiCl.

I.e. 73.89 g of Li2CO3 84.78 g of LiCl.

We can write that,

If         73.89 g of Li2CO3 84.78 g of LiCl

Then   1.5000 g of Li2CO3 say ‘M’ g of LiCl

By cross multiplication method,

M x 73.89 = 1.5000 x 84.78

M = 1.5000 x 84.78/73.89

M = 1.7211 g

From 1.5000 g of Li2CO3 1.7211 g of LiCl will be produced.

==============================XXXXXXXXXXXXXXXXXXX========================

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.