Question

PLEASE SOLVE THIS FOR ME:-
CaCl, CO₂ and H₂O are produced by the reaction of dilute HCI with CaCO₃ . How many grams of HCI will be required to prepare 55.5g of CaCl, 22.2 g of CO₂ and 9.0 g of H₂O from 50.0g of CaCO₃? What is the number of mole of HCI required? [H = 1, Cl = 35.5]​

Answer 1

Answer:

36.7 grams of HCl is required. Also, the number of moles of HCl is nearly 1 .

Explanation:

Let the required mass of HCl be x grams. The skeletal chemical equation for the given reaction is :-

CaCO₃ + HCl → CaCl + CO₂ + H₂O

In this reaction CaCO₃ and HCl are reactants whereas CaCl, CO₂ and H₂O are products. Now, according to "Law of Conservation of mass", we know that :-

Mass of reactants = Mass of products

⇒ 50 + x = 55.5 + 22.2 + 9  

⇒ 50 + x = 86.7

⇒ x = 86.7 - 50  

⇒ x = 36.7 grams

________________________________

Molar mass of HCl is (35.5 + 1) = 36.5 g/mol.

⇒ Moles = Given mass/Molar mass  

⇒ Moles = 36.7/36.5  

⇒ Moles ≈ 1

Hope it helps you :)

Answer 2

Answer:

Using morality, Number of moles are calculated for 1000ml or 1L of the solution.

1000ml contains 0.75MHCl

HCl=0.75 mole.

Therefore, 25ml of 0.75M HCl will contains HCl=

1000

0.75×25

=0.01875mole

2 moles of HCl reacts with =1mole of CaCO 3

Therefore,

0.01875 mole of HCl will react with = 1×0.01875

=0.009375mole

Molar mass of CaCO 3

=100g

Hence, the mass of 0.009375 of CaCO 3

=no.ofmoles×molarmass=0.9375g

You to be happy,

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