Calculate the volume of Carbon dioxide gas generated when 40.0 g of sodium carbonate reacts with hydrocholric acid according to the following reaction. The gas is to be collected at STP
Na2CO3 + HCL-----> NaCL +H2O + CO2(g)
Balance the eqn
Na2CO3 + 2HCl -> 2NaCl +H20 + CO2
From the balanced eqn you can see that one mole of CO2 is produced.
Then at stp 1 mole of gas takes up 24 litres (I think, check this on google) therefore v= 24 litres.
:)
Convert 40 g Na2CO3 to moles. moles = grams/molar mass. Since 1 mole Na2CO3 produces 1 mole CO2, moles Na2CO3 above will contribute that many moles CO2 @ STP. Remember 1 mole CO2 @ STP occupies 22.4 L.
To calculate the volume of carbon dioxide gas generated, you will need to use the ideal gas law equation and the stoichiometry of the reaction.
1. Determine the molar mass of sodium carbonate (Na2CO3) and carbon dioxide (CO2). The molar mass of Na2CO3 is 106 g/mol, and the molar mass of CO2 is 44 g/mol.
2. Convert the given mass of sodium carbonate (40.0 g) to moles. Use the formula:
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
Moles of Na2CO3 = 40.0 g / 106 g/mol
Moles of Na2CO3 ≈ 0.377 mol
3. Use the balanced equation to determine the mole ratio of sodium carbonate to carbon dioxide. From the balanced equation:
1 mol Na2CO3 produces 1 mol CO2
Therefore, 0.377 mol Na2CO3 will produce 0.377 mol CO2.
4. Apply the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP - Standard Temperature and Pressure; 1 atm)
V = volume (unknown)
n = number of moles (0.377 mol CO2, as calculated in step 3)
R = ideal gas constant (0.0821 atm·L/mol·K)
T = temperature (STP - 273.15 K)
Rearranging the equation, we get:
V = (nRT) / P
Plugging in the values:
V = (0.377 mol) * (0.0821 atm·L/mol·K) * (273.15 K) / 1 atm
V ≈ 8.19 L
Therefore, the volume of carbon dioxide gas generated when 40.0 g of sodium carbonate reacts with hydrochloric acid is approximately 8.19 L at STP (Standard Temperature and Pressure).