Question

Barium chloride reacts with sodium sulphate to form barium sulphate and sodium chloride. In an experiment to verify law of conservation of mass 20.8 g of barium chloride reacted with 14.2 g of sodium sulphate to form 11.7 g of sodium chloride. The amount of Barium sulphate formed in the above reaction will be: (a) 23.0 g (b) 23.3 g (c) 23.7 g (d) 35.0 g

Answer 1

b
ac to law of conservation of mass

20.8 + 14.2 = 11.7 +x
x =35 - 11.7
x = 23.3proved

Answer 2

b) 23.3 g. The amount of Barium sulphate formed in the given reaction will be 23.3 g.

Explanation:

Given,

Mass of barium chloride = $20.8 g$

Mass of sodium sulphate = $14.2 g$

Mass of sodium chloride = $11.7 g$

Given reaction

Barium chloride + Sodium sulphate = Barium sulphate + Sodium chloride

$BaCl_2 + Na_2So_4 = BaSo_4 + 2NaCl$

Now, according to the law of conservation of mass we have

$20.8g + 14.2g = BaSo_4 + 11.7g$

$BaSo_4 + 11.7g = 35g$

$BaSo_4 = 35g - 11.7g$

$BaSo_4 = 23.3 g$

Hence, the amount of Barium sulphate formed in the above reaction will be 23.3 g.

Law of Conservation of Mass

According to the law of conservation of mass, matter cannot be created nor destroyed in a closed or isolated system. It is conserved but capable of changing forms. According to the law of conservation of mass, the mass of the reactants and products in a chemical reaction must be equal.

Learn more about the law of conservation of mass

https://brainly.in/question/79592?msp_srt_exp=6

More about the law of conservation of mass

https://brainly.in/question/6246233?msp_srt_exp=6

#SPJ3