Sodium metal reacts with chlorine gas to form NaCl as shown here:
2Na(s) + Cl2(g) -> 2NaCl(s)
If 3.06 L of chlorine gas at STP are completely reacted, how many grams of NaCl will be formed?
Here sodium metal reacts with chlorine as follows
2Na(s) + Cl2(g) -> 2NaCl(s)
So, 2mole of Na reacts with 1mole of Cl2 to give 2mole of NaCl.
Here given that 3.06L of chlorine completely react at STP.
At STP 1 mole of any ideal gas occupies 22.72 L. In other words, the molar volume of a gas at STP, i.e. the volume occupied by 1 mole of gas, is equal to 22.72 L.
Now, 3.06L = 3.06/22.72 mole = 0.1347mole
Hence 0.1347 mole of Cl2 react with 0.2694mole of Na to produce 0.2694mole of NaCl.
Now, molar mass of NaCl = 58.44g
So, 0.2496mole of NaCl = 14.587g
Therefore 14.587 gram of NaCl will be formed.
Thank you
Sodium metal reacts with chlorine gas to form NaCl as shown here: 2Na(s) + Cl2(g) ->...
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