We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.
2) Equation:
2Na OH + H2SO4 --> Na2 SO4 + 2H2O
3) molar ratios
2 mol NaOH : 1 mol H2SO4
4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution
M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4
5) Number of moles of NaOH
2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH
6) Concentration of the solution of NaOH
M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M
7) Standardize the solution of HCl
Chemical reaction:
NaOH + HCl --> NaCl + H2O
8) Molar ratios
1 mol NaOH : 1 mol HCl
9) Number of moles of NaOH in 27.5 ml
M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH
10) Number of moles of HCl
1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl
11) Concentration of the solution of HCl
M = n / V = 0.01169 mol / 0.100 l = 0.1169 M
Rounded to 3 significant figures = 0.117 M
Answers:
[NaOH] = 0.425 M
[HCl] = 0.117 M
