Question

Calcium carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCO3 + 2HCl → CaCl 2 + H2O + CO2 1.00 g of calcium carbonate is added to 50.0 cm3 of 0.0500 mol / dm3 hydrochloric acid. Which volume of carbon dioxide is made in this reaction?

Answer 1

Answer:

.00125 moles of co2 will be formed

Explanation:

Given reaction =  CaCO3 + 2HCl → CaCl 2 + H2O + CO2

amount of species ⇒ CaCO3 = 1.00g

⇒ HCl = 50.0 cm³ of 0.0500 mol/dm³

⇒ Now let us calculate moles of species by formula =

Given mass / molar mass

moles of caco3 = 1/ 100 = 0.01 mol

moles of hcl = 50.0 cm³ = 0.05 dm³

therefore  moles of hcl = 0.05 = x/ 0.05         ∵ molality = moles /volume

x = 0.05*0.05 = .0025 mol

using stoichiometry 1 mol of caco3 required 2 moles of hcl to give 1 mol of co2

∴ only .0025/2 = .00125 moles of co2 will be formed

#SPJ1

Answer 2

Answer:

0.00125 moles of CO₂ will be formed upon the given reaction with the given conditions.

Explanation:

The given reaction is $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$

We are given that 1 g of CaCO₃ reacts with 50 cm³ of HCl

We now calculate the number of moles of HCl by using the formula $no.\;of\;moles = \frac{given\;mass}{molar\;mass}$

$n_C_a_C_O_3 = \frac{1}{100} = 0.01 mol$

$V_H_C_l=50\;cm^3=0.05\;dm^3$

Thus, $n_H_C_l= 0.0025\;mol$

Looking at the given reaction, we understand that 1 mol of CaCO₃ reacts with 2 moles of HCl and produces 1 mole of CO₂.

Thus, with the given stoichiometry, we understand that 0.00125 moles of CO₂ will be formed.

#SPJ1