In a reaction 73 g of Hydrochloric acid reacts with 46 gram of Sodium. The products formed are 2 g of Hydrogen and some Sodium chloride. Calculate the mass of Sodium chloride produced. Which law is applied for the calculation of the mass of Sodium chloride in this reaction?
Answers
Given :-
◉ 73 g of Hydrochloric acid reacts with 46 g of Sodium. The products formed are 2 g of Hydrogen and some sodium chloride.
To Find :-
◉ Mass of sodium chloride
Solution :-
We know, Mass on the reactant side is always equal to the mass on the product side according to the law of conservation of mass. Which states that mass can neither be created nor destroyed.
So, We have the following reaction,
⇒ 2Na + 2HCl ➞ 2NaCl + H₂
It is given that, mass of
- Na = 46 g
- HCl = 73 g
- H₂ = 2 g
According to the law of conservation of mass,
⇒ Mass of reactants = Mass of products
⇒ 46 + 73 = 2 + x [ mass of NaCl be x ]
⇒ x = 71 + 46
⇒ x = 117 g
Hence, The mass of sodium chloride produced is 117 gram.
Alternative :-
The same problem can also be solved using mole concept.
From the reaction we have,
⇒ 2 mole of Na produces 2 mole of NaCl
But the mass of Na is given 46 g, Let us find moles
⇒ no. of mole = mass given / molar mass
⇒ no. of mole = 46 / 23
⇒ no. of mole = 2
Hence, NaCl produced would also be 2 mole
Let's find the mass then,
⇒ no. of mole = mass given / molar mass
⇒ 2 = given mass / 58.5
⇒ mass = 58.5 × 2
⇒ mass = 117 gram.