Question

What is the volume (in litres) of carbon dioxide liberated at STP, when 3 gram of sodium carbonate is treated with excess dilute HCI? (Molecular weight of Na₂CO₃ = 106.

Answer 1

Data Analysis

Na₂CO₃ + 2HCl ―› 2NaCl + H₂O + CO_{2                                                           From the above equation ;}              

–mass of Na₂CO₃= (No. moles × molar mass)  

= 1mol × 106g/mol= 106g    

–volume at standard temperature and pressure(STP)= 22400cm³          –no. moles of CO₂= 1mol                    ------> Volume of CO₂= 1 × 22400cm³= 22400cm³

Solution

So,

*106g of Na₂CO₃―> 22400cm³ of CO₂                                                         *3g of Na₂CO₃―>  x

―> x= 3g × 22400cm³                                           106g                                      –> x=    633.96cm³= 0.63396dm³

Since 1dm³= 1litre, —> 0.63396dm³= 0.63396litre  

Answer 2

Na₂CO₃ + 2HCl → 2Nacl + H₂O + CO₂ 3g

∵ 106 gram Na₂CO₃ liberate = 22.4l CO₂ at STP

∴ 1 gram Na₂CO₃ will liberate CO₂ = \(\frac{22.4}{106}\)L at STP

∴ 3 gram Na₂CO₃ will liberate CO₂ = \(\frac{22.4}{106}\) x 3 L at STP

= 0.634L STP

Hence, volume of CO₂ liberate = 0.63L at STP.

Answer 3

the volume of carbon dioxide liberated at STP when 3 grams of sodium carbonate is treated with excess dilute HCl is approximately 0.634 liters (or 634 milliliters).