An unknown Group 1 metal carbonate, M2CO3, was reacted with excess 2 M HClThe initial mass was 2.002 g of M2CO3and the mass of CO2released was 1.206 g. a) Write a balanced chemical equation for this reaction.

The balanced equation of the reaction is given as follows:

M₂CO₃ + 2HCl ---> 2MCl + H₂O + CO₂

The mole ratio of the metallic carbonate to carbon (iv) oxide is 1:1

The metal is Lithium

Explanation:

The balanced equation of the reaction is given below:

M₂CO₃ + 2HCl ---> 2MCl + H₂O + CO₂

From the equation of reaction, 1 ,mole of the metal carbonate reacts with 2 moles of excess HCl to produce 2 moles of the metallic salt, 1 mole of water and 1 mole of CO₂.

The mole ratio of the metallic carbonate to carbon (iv) oxide is 1:1

Molar mass of CO₂ = 44 g/mol

Number of moles of CO₂ in 1.206 g of CO₂ = 1.206 g / 44 g/mol = 0.0274 moles

Therefore, 2.002 g of M₂CO₃ = 0.0274 moles since all the metallic carbonate was reacted with excess HCl

mass of 1 mole of M₂CO₃ = 2.002/0.0274 = 73.06 g

M₂CO₃ =73.06

2 * M + 12 + 3 *16 = 73.06

2M = 73.06 - 60

M = 13.06/2

M = 6.53 g

Since, the metal belongs to group 1, the metal is most likely Lithium as the molar mass of M is closest to Lithium.