Question

Sodium chloride of 70.2% purity is used to produce salt cake, Na2SO4 of 85.2% purity according to the equation 2NaCl+H2SO4v--> Na2SO4+2HCl Calculate the number of kg of NaCl (impure) required to produce 1 kg of salt cake (impure). Can someone please help me out with this quicky?

Answer 1

Answer:

Explanation:

ImPurity×moles×2=impurity×moles

29.8×moles×2=14.8×7.04

Moles=1.75

Weight=102.3

Answer 2

Sodium Chloride (impure) required to produce 1 kg of salt cake (impure) is 1 kg

Explanation:

The given reaction is

$2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl$

It is a balanced chemical reaction

Here, 2 moles of $NaCl$ produces one mole of $Na_2SO_4$

1 kg of impure salt cake = 1000 gm of impure salt cake

Therefore, the weight of pure salt cake =852 gm

Molecular weight of $Na_2SO_4$ = 142

Therefore, moles of $Na_2SO_4$ = 852/142 = 6

Therefore, moles of $NaCl$ required to produce 4.94 moles of $Na_2SO_4$ = 2 × 6 = 12 moles

Molecular weight of $NaCl$ = 58.5

Weight of $NaCl$ (pure) required to produce $Na_2SO_4$ = 12 × 58.5 gram = 702 gram

Purity of $NaCl$ = 70.2 %

Therefore, $NaCl$ impure required = 702/0.702 gram = 1000 gram = 1 kg

Therefore, 1 kg of $NaCl$ (impure) is required to produce 1 kg of salt cake (impure)

Hope this answer is helpful.

Know More:

Q: 2.34 gram of'impure sodium chloride is treated  with excess of AgNO, solution then 2.87 g of  silver chloride (Mw = 143.5) precipitate is formed.  What is the percentage purity of sodium chloride?

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