Question

Exactly 3.705kg of substance Y are needed to neutralise 100 moles of HCL aq. What could be substance Y ? A. Mgcl2 and Caco3 B. Mgco3 and Ca(NO3)2 C. Mg(NO3)2 and Ca(NO3)2 D. Mgo and Cao

Answer 1

C) Ca(OH)2

Note: The options from the question are wrong. The correct options are given below:

15. Exactly 3.705 kg of substance Y are needed to neutralise 100 moles of HCl(aq).

What could be substance Y?

A Ca B CaO C Ca(OH)2 D CaCO3

Step-by-step explanation:

A) Ca + 2HCl ---> CaCl2 + H2

Molar mass of Ca = 40 g/mol

Mass of Ca required to neutralize 100 moles of HCl = 1/2 * 100 * 40 g = 2000g or 2 kg

B) CaO + 2HCl ---> CaCl2 + H20

Molar mass of CaO = 56.1 g/mol

Mass of CaO required to neutralize 100 moles of HCl = 1/2 * 100 * 56.1 g of CaO = 2805 g or 2.805 kg

C) Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

Molar mass of Ca(OH)2 = 74.1 g/mol

Mass of Ca(OH)2 required to neutralize 100 moles of HCl = 1/2 * 100 * 74.1 g = 3705 g or 3.705 kg

D) CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 required to neutralize 100 moles of HCl = 1/2 * 100 * 100 = 5000 g or 5.00 kg

Therefore, the correct option is C.