An unknown group 1 metal carbonate M2CO3 (M=Li, Na or K) was reacted with excess 2 M HCl and the mass of CO2 released was determined by mass difference

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An unknown group 1 metal carbonate M2CO3 (M=Li, Na or K) was reacted with excess 2 M HCl and the mass of CO2 released was determined by mass difference. The initial mass of solid M2CO3 was 2.002 g and the mass of CO2 released was 1.206 g. a. Write the balanced chemical equation for the reaction of M2CO3 with HCl. b. What is the mole ratio of CO2 to M2CO3 c. Calculate the molar mass of the unknown metal carbonate and identify the group 1 metal.

The Correct Answer and Explanation is:

Let’s break down the problem step by step:

a) Balanced Chemical Equation

When a metal carbonate reacts with hydrochloric acid (HCl), it produces a metal chloride (MCl), carbon dioxide (CO₂), and water (H₂O). For the unknown metal carbonate M₂CO₃, the reaction with excess HCl can be written as:M2CO3(s)+2HCl(aq)→2MCl(aq)+CO2(g)+H2O(l)M_2CO_3 (s) + 2 HCl (aq) \rightarrow 2 MCl (aq) + CO_2 (g) + H_2O (l)M2​CO3​(s)+2HCl(aq)→2MCl(aq)+CO2​(g)+H2​O(l)

This equation shows that one mole of M₂CO₃ reacts with 2 moles of HCl to produce 1 mole of CO₂, 2 moles of MCl, and 1 mole of H₂O.

b) Mole Ratio of CO₂ to M₂CO₃

From the balanced equation, we can observe that the mole ratio between CO₂ and M₂CO₃ is 1:1. That is, 1 mole of M₂CO₃ produces 1 mole of CO₂.

c) Calculate the Molar Mass of the Unknown Metal Carbonate and Identify the Group 1 Metal

  1. Determine the moles of CO₂ released:

We know the mass of CO₂ released (1.206 g). To find the number of moles of CO₂, we use the molar mass of CO₂, which is 44.01 g/mol:moles of CO₂=1.206 g CO₂44.01 g/mol=0.0274 mol CO₂\text{moles of CO₂} = \frac{1.206 \, \text{g CO₂}}{44.01 \, \text{g/mol}} = 0.0274 \, \text{mol CO₂}moles of CO₂=44.01g/mol1.206g CO₂​=0.0274mol CO₂

  1. Use the mole ratio to find moles of M₂CO₃:

From the balanced equation, we know that the mole ratio of CO₂ to M₂CO₃ is 1:1. Therefore, the moles of M₂CO₃ are also 0.0274 mol.

  1. Calculate the molar mass of M₂CO₃:

The mass of M₂CO₃ used in the reaction is 2.002 g. Using the formula for molar mass:Molar mass of M₂CO₃=mass of M₂CO₃moles of M₂CO₃=2.002 g0.0274 mol=73.1 g/mol\text{Molar mass of M₂CO₃} = \frac{\text{mass of M₂CO₃}}{\text{moles of M₂CO₃}} = \frac{2.002 \, \text{g}}{0.0274 \, \text{mol}} = 73.1 \, \text{g/mol}Molar mass of M₂CO₃=moles of M₂CO₃mass of M₂CO₃​=0.0274mol2.002g​=73.1g/mol

  1. Identify the group 1 metal:

The molar mass of M₂CO₃ is 73.1 g/mol. The formula for the molar mass of M₂CO₃ is:M2CO3:2×(molar mass of M)+60.01 g/mol (for CO₃)M_2CO_3: 2 \times \text{(molar mass of M)} + 60.01 \, \text{g/mol (for CO₃)}M2​CO3​:2×(molar mass of M)+60.01g/mol (for CO₃)

So, the molar mass of M (the metal) can be found by subtracting the molar mass of CO₃ (60.01 g/mol) from the total molar mass:Molar mass of M₂CO₃=73.1 g/mol\text{Molar mass of M₂CO₃} = 73.1 \, \text{g/mol}Molar mass of M₂CO₃=73.1g/mol2×(molar mass of M)+60.01=73.12 \times \text{(molar mass of M)} + 60.01 = 73.12×(molar mass of M)+60.01=73.12×(molar mass of M)=73.1−60.01=13.092 \times \text{(molar mass of M)} = 73.1 – 60.01 = 13.092×(molar mass of M)=73.1−60.01=13.09Molar mass of M=13.092=6.545 g/mol\text{Molar mass of M} = \frac{13.09}{2} = 6.545 \, \text{g/mol}Molar mass of M=213.09​=6.545g/mol

The molar mass of the metal is approximately 6.5 g/mol, which corresponds to Lithium (Li), with a molar mass of about 6.94 g/mol.

Final Answer:

  • The balanced equation is:
    M2CO3(s)+2HCl(aq)→2MCl(aq)+CO2(g)+H2O(l)M_2CO_3 (s) + 2 HCl (aq) \rightarrow 2 MCl (aq) + CO_2 (g) + H_2O (l)M2​CO3​(s)+2HCl(aq)→2MCl(aq)+CO2​(g)+H2​O(l)
  • The mole ratio of CO₂ to M₂CO₃ is 1:1.
  • The molar mass of M₂CO₃ is 73.1 g/mol, and the metal in the carbonate is Lithium (Li).

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