Question

Calculate the mass of barium sulphate formed when 500 cm3 of 1 mol dm sodium sulphate solution is reacted completely with barium chloride solution.

[ Relative atomic mass : O = 16; S = 32; Ba = 1371​

Answer 1

116.69 g.

Step-by-step explanation:

Let's write down the reaction taking place when barium chloride reacts with sodium sulfate. This is a double displacement reaction, meaning the anions will be exchanged to produce barium sulfate and sodium chloride:

`BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2 NaCl(aq)`

Let's find the moles of sodium sulfate. Given the volume of:

`V_(Na_2SO_4)=500 cm^(3)`

Convert this to milliliters knowing that:

`1 mL = 1 cm^3`

Therefore:

`V_(Na_2SO_4)=500 cm^3\cdot (1 mL)/(1 cm^3) =500 mL\cdot(1 L)/(1000 mL) =0.500 L`

Also, we know that:

`1 dm^3 = 1 L`

Convert the molarity into mol/L:

`1 (mol)/(dm^3)\cdot (1 dm^3)/(1 L)=1 (mol)/(L)`

Multiply molarity by volume to find moles of sodium sulfate reacted:

`n_(Na_2SO_4)=1 (mol)/(L)\cdot0.500 L=0.500 mol`

According to stoichiometry of the equation, 1 mole of sodium sulfate produces 1 mole of sulfate, so 0.500 mol of sodium sulfate produce 0.500 mol of barium sulfate. That said, we have moles of barium sulfate produced. To find mass, let's multiply this amount of moles by the molar mass of barium sulfate:

`m_(BaSO_4)=0.500 mol\cdot233.38 (g)/(mol) =116.69 g`