Question

0.90 31 gram of mixture of NaCl and KCL on treatment with H2 s o4 yielded 1.0 784 gram of a mixture of N2 s o4 and k 2 s o4 calculate the percentage composition of the mixture...guys plz help​

Answer 1

Explanation:

Solution:

Nacl = X gram and Kcl = 0.9031 - X gram

Nacl + Kcl 》0.9031 g

Na2so4 + K2so4 》 1.0744 gram

Nacl molecular weight = 23 + 35.5 = 58.5 grams

Kcl molecular weight = 39 + 35.5 = 74.5 grams

Now number of moles of Nacl = 1/2 X / 58.5

Number of moles of Kcl =:0.9031 - X / 74.5

so 1/2 X / 58.5 + 1/2 ( 0.9031 - X / 74.5 ) = 1.0784 grams

so X = 0.518 gram

% of Nacl = 0.518 / 0.9031 × 100 = 57.36 %

% of Kcl = 0.3851 / 0.9031 × 100 = 42.64 %