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Question

Three mol of Na2CO3 is allowed to react with a solution containing 6 mol of HCl. Find the volume of CO2 gas (in L) produced at STP.
Na2CO3+2HCl2NaCl+CO2+H2O

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Solution

From the reaction:Na2CO3+ 2HCl2NaCl+CO2+H2OGiven mol:3 mol 6 molGiven mol ratio:1 2Stoichiometric coefficients ratio:1 2

Since given mol ratio of reactants is the same as stoichiometric coefficient ratio, therefore no reactant will be left over.

Now use mol - mol analysis to calculate mol, and hence, the volume of CO2 produced at STP is,
Moles of Na2CO31=Moles of CO2 produced1
Moles of CO2 produced = 3 mol
Volume of CO2 produced at STP = 3×22.4 L=67.2 L

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