Answer:
A. H2SO4 + Ca(OH)2 —> CaSO4 + 2H2O
B. Calcium hydroxide Ca(OH)2 is the limiting reactant.
C. The mole of the unreacted starting material is 0.022 mole.
Explanation:
A. The balanced equation for the reaction.
This is illustrated below:
H2SO4 + Ca(OH)2 —> CaSO4 + 2H2O
B. Determination oh the limiting reactant. The limiting reactant reactant can be obtained as follow:
Step 1:
Determination of the masses of H2SO4 and Ca(OH)2 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 Ca(OH)2 = 40 + 2(16 + 1) = 40 + 2(17) = 40 + 34 = 74g/mol.
Thus, from the balanced equation above,
98g of H2SO4 reacted with 74g of Ca(OH)2.
Step 2:
Determination of the limiting reactant. The limiting reactant can be obtained as follow:
From the balanced equation above,
98g of H2SO4 reacted with 74g of Ca(OH)2.
Therefore, 75g of H2SO4 will react with = (75 x 74)/98 = 56.63g of Ca(OH)2.
From the calculations made above, we discovered that it will take a higher mass of Ca(OH)2 i.e 56.63g than what was given i.e 55g to react with 75g of H2SO4.
Therfore, Ca(OH)2 is the limiting reactant.
C. Determination of the moles of unreacted starting material.
The excess reactant is H2SO4.
The unreacted moles of H2SO4 can be obtained as follow:
Step 1:
Determination of the mass of H2SO4 needed to react with 55g of Ca(OH)2. This is illustrated below:
From the balanced equation above,
98g of H2SO4 reacted with 74g of Ca(OH)2.
Therefore, Xg of H2SO4 will react with 55g of Ca(OH)2 i.e
Xg of H2SO4 = (98 x 55)/74
Xg of H2SO4 = 72.84g
Therefore, 72.84g of H2SO4 reacted.
Step 2:
Determination of the unreacted Mass of H2SO4. This is shown below:
Mass of H2SO4 given = 75g
Mass of H2SO4 that reacted = 72.84g
Mass of unreacted H2SO4 =..?
Mass of unreacted H2SO4 = (Mass of H2SO4 given) – (Mass of H2SO4 that reacted)
Mass of unreacted H2SO4 = 75 – 72.84
Mass of unreacted H2SO4 = 2.16g
Step 3:
Conversion of the unreacted mass of H2SO4 to mole. This is illustrated below:
Mass of unreacted H2SO4 = 2.16g
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98g/mol
Number of mole of unreacted H2SO4 =?
Number of mole = Mass/Molar Mass
Number of mole of unreacted H2SO4 = 2.16/98 = 0.022 mole.
Therefore, the mole of the unreacted starting material is 0.022 mole.
