A solution containing 18.27 g of barium nitrate was added to a solution containing 7.1 g of sodium sulfate.

| November 19, 2020 | education

A solution containing 18.27 g of barium nitrate was added to a solution containing 7.1 g of sodium sulfate. Calculate the mass of the resulting residue.

Na2SO4 + Ba (NO3) 2 = 2 NaNO3 + BaSO4 (sediment)
We recognize the molar masses of the salts by the periodic table:
M (Na2SO4) = 142 g / mol
M (Ba (NO3) 2) = 261g / mol
M (BaSO4) = 233 g / mol
Knowing the molar ratios of these salts according to the reaction equation, we determine which of the reagents is in excess.
n = m / M
n = 7.1 g / 142 g / mol = 0.05 mol – amount of substance Na2SO4
n = 18.27 g / 261 g / mol = 0.07 mol – amount of substance Ba (NO3) 2
This means barium nitrate is taken in excess. Calculations of the sediment mass will be carried out using sodium sulfate.
According to the equation:
n (BaSO4) = n (Na2SO4) = 0.05 mol
m = n * M
m = 0.05 mol * 233 g / mol = 11.7 g. is the mass of barium sulfate.
Answer: 11.7 g is the mass of the resulting residue.



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