Answer:
d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.
Explanation:
From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)
2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3
Hence 1 mole of Br combines with 2/3 moles of Al
or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving
0.4 - 0.267 = 0.133 moles of Al remaining unreacted
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol
Molarity = number of mole of substance(n) / volume of solution (V).
n(CaCl2) = mass (CaCl2)/M(CaCl2)
M(CaCl2) = 40+2*35.5 = 111 g/mol
n(CaCl2) =39.5 g CaCl2*1 mol/111g
0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).
volume of solution (V) = 39.5 g CaCl2*1 mol/(0.250 M*111g) = 1.42 L
Answer : The results would show more amount of water in the hydrated sample.
Explanation :
The amount of water of crystallization can be found by taking the masses of hydrated copper sulfate and anhydrous copper sulfate.
The difference in masses indicates the mass of water lost during dehydration process.
If during dehydration process, some of the copper sulfate spatters out of the crucible, then this would give us less mass for anhydrous sample than the actual.
As a result, the difference in masses of hydrated sample and the anhydrous sample would be more.
Therefore the results would show more amount of water in the hydrated sample.
