Question

2.05 g of the carbonate of an unknown alkali metal (X2CO3) required 8.9 cm3 of 2.0 mol dm-3 hydrochloric acid to completely dissolve it. What was the relative atomic mass of the metal and which metal was it? Equation: X2CO3(s) + 2HCl(aq) ....> 2XCl(aq) + CO2(g) + H2O(l)

Answer 1

Answer:

Na2co3+2Hcl(aq)-->2cl(aq)+CO2(g)+H20(l)

Answer 2

Answer:

The relative atomic mass of the metal present in the carbonate compound which dissolved in 8.9 cm³ of 2.0 mol dm⁻³ HCl is 85 and the metal is Rubidium ( Rb)

Explanation:

X₂CO₃ + 2HCl   →  2XCl + CO₂ + H₂O

So, 2 moles of HCl reacts with 1 mole of X₂CO₃.

given that,

              volume of HCl  = 8.9 cm³ = 8.9×10⁻³ dm³

    concentration of HCl =2.0 mol dm⁻³

                 Moles of HCl =volume (dm³) x  concentration (mol dm⁻³)

                                        = 8.9×10⁻³×2.0 = 17.8 ×10⁻³  mol

Thus, Moles of X₂CO₃    = $\frac{1}{2}$ × Moles of HCl

                                        =$\frac{1}{2}$ ×17.8 ×10⁻³ = 8.9×10⁻³ mol

Given that,

mass of X₂CO₃ = 2.05 g

molecular mass of X₂CO₃ =mass of X₂CO₃ ÷ Moles of X₂CO₃

                                           =2.05 / 8.9×10⁻³ = 230.33 ≈ 230 g/mol

molar mass of carbonate =  60 g/mol

then, atomic mass of metal given by,

          2x +60 = 230

                 x   =(230-60)/2 = 85

The relative atomic mass of metal is 85 u. Hence the alkali metal is Rubidium( Rb).