Answer to Question #255301 in General Chemistry for Jaysin

Question

How many milliliters of dry CO₂, measured at STP, could be evolved in the reaction between 17.0 mL of 0.293 M NaHCO₃ and 31.0 mL of 0.864 M HCl?

Answer 1

First find out the mmoles of NaHCO₃and HCl

moles = Molarity × volume

moles of NaHCO₃= ( 0.270M ×16.2mL ) = 4.374 moles

moles of HCl = ( 0.753M × 34.2mL ) = 25.75 moles

According to the chemical reaction equation,

1mol of NaHCO₃ reacts with 1mol of HCl to Produces 1mol of CO₂

Then,

4.374mol of NaHCO₃ react with 4.374mol of HCl to Produces 4.374mol of CO₂

Calculate the volume of CO₂ :

At STP, molar volume = 22.4mol/L

moles of CO₂ = $\frac{ volume}{molar volume}$

4.374mol = $\frac{volume}{22.4mol/L}$

Volume = 4.374mol ×22.4mol/L

Volume = 98mL

Volume of CO₂ at STP, is 98mL