Answer:
0.72g
Explanation:
Step 1:
We'll begin by writing a balanced equation for the reaction. This is illustrated below:
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Step 2:
Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol
Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Step 3
Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.
Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:
From the balanced equation above,
98g of H2SO4 required 80g of NaOH.
Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e
Xg of H2SO4 = (98x1.6)/80
Xg of H2SO4 = 1.96g
Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.
Step 4:
Determination of the mass of water produced from the reaction. This is illustrated below:
The balanced equation for the reaction is given below:
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g
From the balanced equation above,
80g of NaOH reacted to produced 36g of H2O.
Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.
Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g
