Question

when 2 moles of Na2CO3 reacts with 6 moles of HCl, the volume of co2 gas liberated at S.T.P will be ?​

Answer 1

Answer:

Explanation:

The Reaction will be Na₂CO₃+2HCl→2NaCl+H₂O+CO₂

Here Na₂CO₃ and HCl are in 1:2 ratio. so 2 mole Na₂CO₃ will only act with 4 moles of HCl.

So Na₂CO₃ IS THE LIMITING REAGENT.

Now, Na₂CO₃ AND CO₂ are in ratio 1:1 . \

so CO₂ produced will be 2 moles.

IN STP THE VOLUME WILL BE 44.8L OR 44800mL

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Answer 2

Given : Number of moles of sodium carbonate - 2

Number of moles of HCl - 6

Find : Volume of CO2 gas liberated by the given reaction at STP.

Solution : The reaction of sodium carbonate with hydrogen chloride is as follows -

Na2CO3 + 2HCl --> 2NaCl + H2O + CO2.

In the reaction, 1 mole of Na2CO3 reacts with 2 moles of HCl. So, 2 moles of Na2CO3 will react with : 4 moles of HCl.

Here HCl is present in excess amount and Na2CO3 is the limiting reagent. Hence, CO2 amount will be determined by the concentration of Na2CO3.

1 mole of Na2CO3 releases 1 mole of CO2, so 2 moles will release 2 moles of CO2.

At STP conditions, 2 moles of CO2 will be : 2*22.4 litre = 44.8 litres.

Thus, volume of CO2 gas liberated will be 44.8 litres.