Question

If 35 grams of solid sodium carbonate are put into 154 mL of a 3.80 M HCl solution, how many liters of carbon dioxide gas would be released at STP? (Answer is: 6.47L)

Answer 1

no of moles of Na2CO3 = 35/ Molar mass Na2CO3 = 35 / 106 = 0.331 moles

no of moles of HCl = M x V ( in L) = 3.80 x 154 x 10-3 = 0.585 moles

Na2CO3 + 2 HCl    $\rightarrow$ 2NaCl + H2O + CO2

0.331 0.585 moles

HCl will be the limiting reagent

2 mole HCl react with 1 mole CO2

so o.585 moles HCl react with 0.585/2 mole CO2

now 1 mole gas at STP equals 22.4 L

so 0.585/2 mole gas at STP equals = 6.5 L  Ans