Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) →  CaCl2(aq) + CO2 (g) + H2O(l). What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Asked by Abhisek | 1 year ago |  143

1 Answer

Solution :-

0.75 M of HCl

≡ 0.75 mol of HCl are present in 1 L of water

≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

\(\dfrac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL\)

= 0.6844 g

Given chemical reaction,

\(CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l) \)

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of \(CaCO_{ 3 }(100 g)\)

Therefore, amt of \(CaCO_{ 3 }\) that will react with 0.6844 g

\(\dfrac{ 100 }{ 73 } \; \times \; 0.6844 \; g\)

= 0.9375 g

Answered by Pragya Singh | 1 year ago

Related Questions

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:

4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Class 11 Chemistry Some Basic Concepts of Chemistry View Answer

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.  Find:

(i) Empirical formula

(ii) Molar mass of the gas, and

(iii) Molecular formula

Class 11 Chemistry Some Basic Concepts of Chemistry View Answer

Calculate the number of atoms in each of the following

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He

Class 11 Chemistry Some Basic Concepts of Chemistry View Answer

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope Molar mass Abundance
\(\, _{ 36 }\textrm{Ar}\) \( 35.96755 g \; mol^{ -1 }\) 0.337 %
\(\, _{ 38 }\textrm{Ar}\) \( 37.96272 g \; mol^{ -1 }\) 0.063 %
\(\, _{ 40 }\textrm{Ar}\) \( 39.9624g \; mol^{ -1 }\) 99.600 %

Class 11 Chemistry Some Basic Concepts of Chemistry View Answer

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope Molar mass Abundance
\(\, _{ 36 }\textrm{Ar}\) \( 35.96755 g \; mol^{ -1 }\) 0.337 %
\(\, _{ 38 }\textrm{Ar}\) \( 37.96272 g \; mol^{ -1 }\) 0.063 %
\(\, _{ 40 }\textrm{Ar}\) \( 39.9624g \; mol^{ -1 }\) 99.600 %

Class 11 Chemistry Some Basic Concepts of Chemistry View Answer