M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.
Answer:
Explanation:
2 H₂(g) + O₂(g) → 2 H₂O(g
2 moles 1 mole 2 mole
50 g of H₂ = 50 /2 = 25 moles of H₂
100 g of O₂ = 100 / 32 = 3.125 moles of O₂
So oxygen is the limiting reagent .
3.125 moles of O₂ will react with 6.25 moles of H₂ to give 6.25 moles of H₂O .
Hence moles of H₂O produced = 6.25 moles .
D. toxic chemical used to control pest population.
<span>Alkanes are unreactive except in combustion reactions.</span>
Answer:

Explanation:
1. Molar concentration
Let's call chloroform C and acetone A.
Molar concentration of C = Moles of C/Litres of solution
(a) Moles of C
Assume 0.187 mol of C.
That takes care of that.
(b) Litres of solution
Then we have 0.813 mol of A.
(i) Mass of each component

(ii) Volume of each component

(iii) Volume of solution
If there is no change of volume on mixing.
V = 15.08 mL + 59.70 mL = 74.78 mL
(c) Molar concentration of C

2. Molal concentration of C
Molal concentration = moles of solute/kilograms of solvent
Moles of C = 0.187 mol
Mass of A = 47.22 g = 0.047 22 kg
