Question

When 8.4 g of nahco3 is added to a solution of ch3cooh weighing 20 g . it is observed that 4.4 g of co2 is released into the atmosphere & a residue is left behind. calculate the mass of residue by applying law of conservation of mass?

Answer 1

The equation for the reaction between the sodium bicarbonate and acetic acid is given below:

NaHCO₃ + CH₃COOH = CH₃COONa + H₂O + CO₂

The law of conservation of mass states that the mass of the reactants is equal to the mass of the products.

Mass of reactants = 20 + 8.4 = 28.4g
Mass of CO₂ produced = 4.4 g

28.4 - 4.4 = 24g

That means mass of the water and sodium acetate formed is 24 g

Use mole ratio to find the mass of the sodium acetate formed

Mole ratio in the equation is 1:1:1:1:1

Find moles of sodium 8.4 g bicarbonate

moles = mass/molar mass
molar mass = 84

moles= 8.4/84
=0.1 moles
Since ratio is 1:1, then the moles of the H₂O is also 0.1

Find mass of water formed

Mass = moles  × molar mass
         = 0.1 × 18
         = 1.8 g

Mass of sodium acetate

mass = moles × molar mass
          = 0.1 × 82
          = 8.2 g

In this equation the sodium bicarbonate is a limiting reagent hence, it will still be present in the reactants.

Answer 2

Answer:

the answer is 24 bcause 28.4-4.4

Explanation: