Question

at STP how much volume of Co2 will obtained when 6 moles each of N A 2 c o 3 and HCL were taken

can someone please answer​

Answer 1

Answer:

1.06 g of sodium carbonate correspond to  1061.06=0.01 moles.

Na2CO3+2HCl→2NaCl+H2O+CO2

1 mole of sodium carbonate liberates 1 mole of CO2.

0.01 moles of sodium carbonate will liberate 0.01 moles of CO2.

1 mole of CO2 at STP occupies a volume of 22400 cm3.

0.01 mole of CO2 at STP will occupy a volume of 224 cm3.