Question

Calcium carbonate found in limestone and marble reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water. what mass of hydrochloric acid will be needed to produce 5.00x103 kg of calcium chloride

Answer 1

Answer : The correct answer is 1.64 * 10³ Kg.

Given : Mass of CaCl₂ = 5.00 * 10³ Kg

Converting Kg to g using ( 1 Kg = 1000 g)

`Mass of CaCl_2 = 5.00 * 10^3 Kg *(1000 g)/(1 Kg )`

Mass of CaCl₂ = 5.00 * 10⁶ g

Following steps can be done :

Step 1) To write a balanced reaction :

The reaction between Calcium Carbonate and Hydrochloric acid as:

`CaCO_3 + 2 HCl \rightarrow CaCl_2 + CO_2 + H_2O`

Step 2 ) To find mole of CaCl₂ .

Mass of compound can be used to calculate mole by Mole formula as :

`Mole (mol) = (given mass (g) )/(molar mass (g)/(mol))`

Mass of CaCl₂ = 5 * 10⁶ g Molar mass of CaCl₂ = 110.98
`(g)/(mol)`

Plugging values in formula

`Mole = (5*10^6 g)/(110.98 (g)/(mol))`

Mole of CaCl₂ = 4.5 * 10^4 mol

Step 3) To write mole ratio of HCl : CaCl₂

Mole ratio is found from balanced reaction .

Mole of HCl in balanced reaction = 2 mol

Mole of CaCl₂ in balanced reaction = 1 mol

hence mole ratio of HCl : CaCl₂ = 2 : 1

Step 4) To find mole of HCl

Mole of HCl can be calculated using mole of CaCl₂ and mole ratio as :

`Mole of HCl = mole of CaCl_2 * Mole ratio`

`Mole of HCl = 4.5 * 10^4 mol * (2)/(1)`

Mole of HCl = 9.0 * 10⁴ mol

Step 5) To find mass of HCl

Mass of HCl can be calculates using its mole by Mole formula as:

Molar mass of HCl = 36.5
`(g)/(mol)`

Mole of HCl = 9.0 * 10⁴ mol

Plugging values in Mole formula as:

`9.0 * 10^4 mol = (mass of HCl )/(36.5 (g)/(mol))`

Multiplying both side by
`36.5 (g)/(mol)`

`9.0 * 10^4 mol * 36.5 (g)/(mol)= (mass of HCl )/(36.5 (g)/(mol)) * 36.5 (g)/(mol)`

Mass of HCl = 3.285 * 10⁶ g

Step 6) Convert mass of HCl from g to Kg

1 Kg = 1000 g

`Mass of HCl = 3.285 *10^6 g *(1 Kg)/(1000 g)`

Mass of HCl = 3.285 * 10³ Kg

Hence mass of HCl required to produce 5.00 * 10³ Kg of CaCl₂ is 3.285 *10³Kg.

Answer 2

The equation for the reaction is;
CaCO3(s) +2 HCl(aq) = CaCl2(aq) + CO2(s) + H2O(l)
Mass of calcium chloride is 5 ×10^3 kg
Moles of calcium chloride will be; (5× 10^6 g)/ 110.98 g/mol = 4.5 × 10^4 moles
Using the mole ratio; HCl : CaCO3 = 2:1
Thus; the number of moles of HCl will be ; 9.0 ×10^4 moles
The mass of HCl = 9.0 × 10^4 moles ×36.46
= 3.29 × 10^6 kg or 3.29 × 10^3 kg