when 8.4 g of NaHCO3 is added to a solution of CH3COOH weighing 20 g .It is observrd that 4.4 g of CO2 is releasd into atmosphere and a residue is left behind .calculate the mass of residue by applying law of conservation of mass .
NaHCO3 + CH3COOH=>CH3COONa + H2O + CO2
8.4........20g.........x......y....4.4
So you start with 20+8.4 = 28.4
You lose 4.4 which should leave you with 24.0 grams. (In case you're interested, that's the CH3COONa(that's x = 8.2 g) + H2O(that's y=1.8 g) + whatever amount the unused acetic acid (that's 14.0 g) and
14.0 + 8.2 + 1.8 = 24.0 g.
To calculate the mass of the residue left behind by applying the law of conservation of mass, we need to consider the reaction that occurs between NaHCO3 (sodium bicarbonate) and CH3COOH (acetic acid). The balanced chemical equation for the reaction is as follows:
NaHCO3 + CH3COOH -> CO2 + H2O + NaCH3COO
From the equation, we can see that 1 mole of NaHCO3 reacts with 1 mole of CH3COOH to produce 1 mole of CO2, along with other products.
1. Calculate the number of moles of NaHCO3:
We are given the mass of NaHCO3, which is 8.4 g. The molar mass of NaHCO3 is:
23.0 (Na) + 1.0 (H) + 12.0 (C) + 3 * 16.0 (O) = 84.0 g/mol
Number of moles of NaHCO3 = mass / molar mass
= 8.4 g / 84.0 g/mol
= 0.1 mol
2. Calculate the number of moles of CO2 produced:
From the balanced equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2. Therefore, 0.1 mol of NaHCO3 will produce 0.1 mol of CO2.
3. Calculate the mass of CO2 produced:
We are given that 4.4 g of CO2 is released. The molar mass of CO2 is:
12.0 (C) + 2 * 16.0 (O) = 44.0 g/mol
Mass of CO2 = number of moles x molar mass
= 0.1 mol x 44.0 g/mol
= 4.4 g
4. Calculate the mass of the residue:
The total mass of the reactants is the sum of the mass of NaHCO3 and CH3COOH, which is given as 20 g.
Mass of residue = Mass of reactants - Mass of CO2 produced
= 20 g - 4.4 g
= 15.6 g
Therefore, the mass of the residue left behind is 15.6 g.
To calculate the mass of the residue using the law of conservation of mass, we need to account for all the reactants and products involved in the reaction.
Given information:
Mass of NaHCO3 added = 8.4 g
Mass of CH3COOH = 20 g
Mass of CO2 released = 4.4 g
The balanced chemical equation for the reaction between NaHCO3 and CH3COOH is:
NaHCO3 + CH3COOH -> CO2 + H2O + NaCH3COO
From the equation, we can see that the reaction produces CO2, water (H2O), and sodium acetate (NaCH3COO).
1. Calculate the molar mass of NaHCO3:
The molar mass of Na = 22.99 g/mol
The molar mass of H = 1.01 g/mol
The molar mass of C = 12.01 g/mol
The molar mass of O = 16.00 g/mol
Now, we calculate the molar mass of NaHCO3:
(1 * 22.99 g/mol) + (1 * 1.01 g/mol) + (1 * 12.01 g/mol) + (3 * 16.00 g/mol) = 84.01 g/mol
2. Determine the number of moles of NaHCO3:
Number of moles = mass / molar mass
Number of moles = 8.4 g / 84.01 g/mol ≈ 0.1 mol
3. Determine the number of moles of CH3COOH:
Number of moles = mass / molar mass
Number of moles = 20 g / (molar mass of CH3COOH)
Note: You need to know the molar mass of CH3COOH to calculate its moles.
4. Convert moles of NaHCO3 to moles of CO2:
From the balanced equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.
5. Determine the mass of residue:
The law of conservation of mass states that the total mass of products obtained must equal the total mass of reactants consumed.
Mass of residue = (mass of CH3COOH) - (mass of CO2 released)
Mass of residue = 20 g - 4.4 g = 15.6 g
Therefore, the mass of the residue is 15.6 g.