Answer:
The answer to your question is:
1.- CO
2.- 0.414 moles of CO2
Explanation:
Data
2CO + O2 ⇒ 2CO2
CO = 0.414 moles
O2 = 0.418
Process
theoretical ratio CO/O2 = 2/1 = 1
experimental ratio CO/O2 = 0.414/0.418 = 0.99
Then the limiting reactant is CO
2.-
2 moles of CO --------------- 2 moles of CO2
0.414 moles of CO --------- x
x = (0.414 x 2) / 2
x = 0.414 moles of CO2
Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
The correct answer would be the last one.
Answer:
It’s twice as much as 9+10
Explanation:
