Question

the weight of Barium Chloride that would react with 24.4 gram of sodium sulphate to produce 46.6 gram of Barium sulphate and 23.4 gram of sodium chloride is​

Answer 1

Answer:

41.64g can be the correct answer

Explanation:

There are few concepts needed for this type of questions which are directly based on Moles.

These concepts are:-

C1: Use must make a balanced equn. for the reaction:-

BaCl2 + Na2SO4 -----> BaSO4 + 2 NaCl

C2: u need to know how many moles of the reactant produce how many moles of product.

This can be identified by seeing the BALANCED CHEMICAL EQUATION as above.

Here, the ratio is 1:1 gives 1:2.

it means 1 mole of BaCl2 and 1 mole of Na2SO4 produces 1 mole of BaSO4 and 2 mole of NaCl.

C3: u shud know how to calculate no. of moles

n = Weight taken/ Molar Mass.

Here n= no. of moles

So, n(BaCl2) =?

n(Na2SO4) = 24.4/ 142.04 ~ 0.2 moles

also, n(BaSO4) = 46.6/ 233.38 ~ 0.2 moles

and, n(NaCl) = 23.4 / 58.44 ~ 0.40 moles

C4: this concept is used to determine the weight or mass taken by using moles.

As we know from C2.

n(BaCl2) = n(Na2SO4) = n(BaSO4) = n(NaCl)/2 ~ 0.2 moles.

now using C3. weight = n× molar mass

====> Weight BaCl2 = 0.2× 208.23 = 41.64g

Hence req. Mass of Barium Chloride is 41.64g

Answer 2

Answer:

45.6

Explanation:

use the law of conservation of mass.

In a chemical reaction, the total mass of the reactants equals the total mass of the products.

i.e. 46.6g+23.4g=24.4g+x

=70-24.4= 45.6g