Question

You treat 9.540 g of the mixture with the acid and isolate 9.355 g of nacl. what is the weight percent of each substance in the mixture?

Answer 1

The weight percentage of sodium carbonate in the mixture is = 67.71%  

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

What does "molar mass" ?

The total mass throughout grams of all the atoms needed to form a molecule per mole is what makes up the molar mass, also known as the molecular weight. Grams per mole is the unit measuring molar mass.

According to the given information:

The interaction between sodium carbonate and hydrochloric acid has the following equation:

Na₂CO₃ + HCl  --> NaCl  + CO₂ + H₂O

The reaction's balanced equation is,

Na₂CO₃ + 2HCl  --> 2NaCl  + CO₂ + H₂O -------(2).

When sodium hydrogen carbonate as well as hydrochloric acid react, the following equation results.

NaCO₃ + HCl  --> NaCl  + CO₂ + H₂O ----------------(3)

The Molar mass :

Na₂CO₃ = 106g/mol

NaCO₃ = 84g/mol

NaCl  = 58.5g/mol

The mass of the mixture is Na₂CO₃/NaCO₃ = 9.540g

The molar masses of the constituent elements that make up the compounds are added to determine the molar weight of the substances. In both chemical equations (2) and (3), the unitary technique is employed to calculate the mass of the mixture based on the number of moles (3).

The mass of the sodium carbonate and sodium bicarbonate is calculated with the help of the mass of NaCl formed by the mixture of  Na₂CO₃/NaCO₃ with HCl.

The mixture has the following percentage of sodium carbonate:

        % of Na₂CO₃ =  $\frac{x \times 106}{9.540} \times 100$

                             $=\frac{0.061 \times 106}{9.540} \times 100$

                             = (6.46/9.540)*100

                             = 67.71%        

The weight percentage of sodium carbonate in the mixture is = 67.71%      

The mixture has the following percentage of NaHCO3:

        % of NaHCO₃ =  $\frac{y \times 84}{9.540} \times 100$

                             = $=\frac{0.037 \times 84}{9.540} \times 100$

                              = (3.108/9.540)*100

                             = 32.57%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

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I understand that the question you are looking for is:

A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. the unbalanced equations for the resulting's reactions are:

Na2CO3 (s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

You treat 9.540g of Na2CO3/NaHCO3 mixture with an excess of aqueous HCl and isolate 9.355g of NaCl. What is the weight percent of each substance in the mixture?