Na2CO3 reacts with H2SO4 to give respective salt water and carbon dioxide. Calculate the mass of pure salt formed when 300 g of Na2CO3 of 80% purity reacts with H2SO4
To calculate the mass of pure salt formed, we need to first determine the amount of Na2CO3 that reacts with H2SO4. Then we can find the mass of pure salt formed using the stoichiometry of the balanced chemical equation.
First, let's calculate the amount of Na2CO3 in grams that reacts with H2SO4.
Given:
Mass of Na2CO3 = 300 g
Purity of Na2CO3 = 80% (80/100 = 0.8)
To find the mass of pure Na2CO3, we multiply the mass of Na2CO3 by its purity:
Mass of pure Na2CO3 = Mass of Na2CO3 * Purity
Mass of pure Na2CO3 = 300 g * 0.8
Mass of pure Na2CO3 = 240 g
Next, let's balance the chemical equation:
Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2
From the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and Na2SO4 is 1:1.
Since the molar mass of Na2CO3 is 105.99 g/mol, we can calculate the amount of Na2CO3 in moles:
Amount of Na2CO3 = Mass of pure Na2CO3 / Molar mass of Na2CO3
Amount of Na2CO3 = 240 g / 105.99 g/mol
Amount of Na2CO3 = 2.26 mol
According to the stoichiometric ratio, the amount of Na2CO3 is the same as the amount of Na2SO4 that will be formed.
Now, we can calculate the mass of pure Na2SO4 formed using its molar mass.
Molar mass of Na2SO4 = 142.04 g/mol
Mass of pure Na2SO4 formed = Amount of Na2SO4 * Molar mass of Na2SO4
Mass of pure Na2SO4 formed = 2.26 mol * 142.04 g/mol
Mass of pure Na2SO4 formed = 321.05 g
Therefore, the mass of pure salt formed when 300 g of 80% pure Na2CO3 reacts with H2SO4 is 321.05 grams.
You need a comma in the question between salt and water. It is salt, water and CO2 (not salt water and CO2).
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
mols Na2CO3 if pure = grams/molar mass = 300/142 = about 2 but that's an estimate.
It's only 80% so 2 x 0.80 = about 1.6 g Na2SO4 actually there.
Convert mols Na2CO3 to mols Na2SO4. Then
g Na2SO4 = mols Na2SO4 x molar mass Na2CO3