Answer:
58.6 % by mass of Na₂CO₃
Explanation:
This is the reaction:
Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O
Let's find out the moles of CO₂ produced, by the Ideal Gases Law
1.24 atm . 1.67 L = n . 0.082 . 299K
(1.24 atm . 1.67 L / 0.082 . 299K) = n
0.0844 moles = n
Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:
2 moles of CO₂ were produced by 1 mol of Na₂CO₃
Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.
Let's convert this moles into mass (mol . molar mass)
0.0422 mol . 106 g/mol = 4.47 g
Finally we can know the mass percent of sodium carbonate in the mixture
(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %
The percent composition by mass of Na₂CO₃ in the mixture is 58.6%
Firstly, write the chemical reaction involved in this:
Na₂CO₃+MgCO₃+4HCl → 2NaCl+2CO₂+2H₂O
We need to find the number of moles of CO₂.
Number of moles (n)=[tex]\frac{1.24 atm * 1.67 L}{0.082 *299K}[/tex]
n=0.0844 moles
So, 1 mol of sodium carbonate produces 2 moles of carbon dioxide.
OR
2 moles of CO₂ were produced by 1 mol of Na₂CO₃.
Then, 0.0844 moles of CO₂ will be= [tex]\frac{1}{2}[/tex]×0.0844=0.0422 moles of Na₂CO₃
In order to get the answer in mass, we need to convert moles into molar mass.
Molar mass of Na₂CO₃= 106g/mol
Molar mass for 0.0422 mol of Na₂CO₃=0.0422×106=4.47 g
In order to find the mass percentage which is given by:
Mass percentage= Mass of Compound/ Total Mass *100
=(4.47g / 7.63g)*100= 58.6%
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