Cu + S ---> CuS
by reaction 1 mol 1 mol
from the problem 0.25 mol 0.25 mol
0.25 mol Cu
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
