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

| March 3, 2021 | 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 from 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

So 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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