Calculate the volume of CO2 produced at S.t.p when 25cm3 of 0.25mol/dm3 Hcl reacts with Na2Co3.
balance the equation:
Na2CO3 +2 HCl >>> 2NaCl+ H2O + CO2
moles HCl: .25*25/1000=6.25e-3 moles
so according to the equation above, it will react with half that sodium carbonate.
Moles Na2CO3; volume/22.4 = 1/2 * .625e-4
volume=22.4*.5*.625e-e dm^3
Don't understand
To calculate the volume of CO2 produced at Standard Temperature and Pressure (STP) when 25 cm3 of 0.25 mol/dm3 HCl reacts with Na2CO3, we need to follow these steps:
Step 1: Write a balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between HCl and Na2CO3 is:
2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2
Step 2: Calculate the number of moles of HCl used.
Given that the concentration of HCl is 0.25 mol/dm3 and the volume used is 25 cm3, we need to convert the volume to dm3:
25 cm3 = 25/1000 dm3 = 0.025 dm3
The number of moles of HCl used can be calculated using the formula:
moles = concentration (mol/dm3) × volume (dm3)
moles = 0.25 mol/dm3 × 0.025 dm3 = 0.00625 mol HCl
Step 3: Use the stoichiometry of the balanced equation to determine the number of moles of CO2 produced.
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of CO2. Therefore, the number of moles of CO2 produced is also 0.00625 mol.
Step 4: Calculate the volume of CO2 produced at STP.
At STP, 1 mole of any ideal gas occupies 22.4 dm3.
Therefore, the volume of CO2 produced at STP can be calculated using the formula:
volume = number of moles × molar volume
volume = 0.00625 mol × 22.4 dm3/mol = 0.14 dm3
So, the volume of CO2 produced at STP when 25 cm3 of 0.25 mol/dm3 HCl reacts with Na2CO3 is 0.14 dm3.