P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present
according to molar ratio of 1:4
number of PCl₅ moles formed -0.22 mol x 4 = 0.88 mol
0.88 mol of PCl₅ is formed
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
Answer:- C. H
Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.
As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.
The oxidation number in elemental form is zero.
In
, the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in
is -1. On product side, the oxidation number of hydrogen in
is zero and in
the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in
is 0.
From above data, Oxidation number of O is -2 on both sides so it is not reduced.
Oxidation number of Cl is changing from -1 to 0 which is oxidation.
Oxidation number of H is changing from +1 to 0 which is reduction.
So, the right choice is C.H
Answer:
Pb(NO3)2 + KI = KNO3 + PbI2