To find the mass of CO2 produced in the reaction, we need to determine the molar mass of CO2 and then calculate the moles of CO2 produced.
The molar mass of CO2 is calculated by adding the molar masses of carbon (C) and oxygen (O):
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 2 atoms x 16.00 g/mol = 32.00 g/mol
Molar mass of CO2 = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
From the balanced equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.
Therefore, to find the moles of CO2 produced, we need to convert the mass of NaHCO3 given to moles.
Mass of NaHCO3 = 24.5 g
Moles of NaHCO3 = (24.5 g)/(84.01 g/mol) = 0.29 mol
As 1 mole of NaHCO3 produces 1 mole of CO2, the moles of CO2 produced in the reaction is also 0.29 mol.
To find the mass of CO2, we can use the formula:
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= 0.29 mol x 44.01 g/mol
= 12.75 g
Therefore, the mass of CO2 produced in the reaction is 12.75 grams.
Question at position 5
What is the mass, in grams, of the CO2 produced in the reaction?
2NaHCO3 → Na2CO3 + H2O + CO2
24. 5g 19.2 g 4.5 g ?
3 answers
no its not
I apologize for any confusion. Let's correct the calculation.
From the balanced equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.
The molar mass of NaHCO3 can be calculated as follows:
Na: 1 atom x 22.99 g/mol = 22.99 g/mol
H: 1 atom x 1.01 g/mol = 1.01 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 48.00 g/mol = 84.01 g/mol
From the given reaction, we have:
2NaHCO3 → Na2CO3 + H2O + CO2
So, for every 2 moles of NaHCO3, we obtain 1 mole of CO2.
Given that the mass of NaHCO3 is given as 24.5 g, we can calculate the moles of NaHCO3:
Moles of NaHCO3 = (24.5 g) / (84.01 g/mol) = 0.2917 mol
Since 2 moles of NaHCO3 produce 1 mole of CO2, we have:
Moles of CO2 = (0.2917 mol) / (2 mol) = 0.1458 mol
To find the mass of CO2, we can use the formula:
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= 0.1458 mol x 44.01 g/mol
= 6.42 g
Therefore, the mass of CO2 produced in the reaction is approximately 6.42 grams.
From the balanced equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.
The molar mass of NaHCO3 can be calculated as follows:
Na: 1 atom x 22.99 g/mol = 22.99 g/mol
H: 1 atom x 1.01 g/mol = 1.01 g/mol
C: 1 atom x 12.01 g/mol = 12.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 48.00 g/mol = 84.01 g/mol
From the given reaction, we have:
2NaHCO3 → Na2CO3 + H2O + CO2
So, for every 2 moles of NaHCO3, we obtain 1 mole of CO2.
Given that the mass of NaHCO3 is given as 24.5 g, we can calculate the moles of NaHCO3:
Moles of NaHCO3 = (24.5 g) / (84.01 g/mol) = 0.2917 mol
Since 2 moles of NaHCO3 produce 1 mole of CO2, we have:
Moles of CO2 = (0.2917 mol) / (2 mol) = 0.1458 mol
To find the mass of CO2, we can use the formula:
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= 0.1458 mol x 44.01 g/mol
= 6.42 g
Therefore, the mass of CO2 produced in the reaction is approximately 6.42 grams.
3 answers