To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
1. Atomic Mass
2. Highest melting point
3. Configuration
4. A
5. A
6. D
7. A
8. C
9. B
10. A
11. C
12. C
13. A
Answer:
0,72 moles of SO₂ remain
Explanation:
The reaction is:
2SO₂ + O₂ → 2SO₃
Where molecular mass of SO₂ is 64,066g/mol and of SO₃ is 80,066g/mol.
86,0g of SO₂ are:
86,0g × (1mol / 64,066g) = <em>1,34 moles of SO₂</em>.
50,0g of SO₃ are:
50,0g × (1mol / 80,066g) = <em>0,62 moles of SO₃</em>.
Now, as 2 moles of SO₂ produce 2 moles of SO₃, the moles of SO₂ that remain after the reaction are the initial moles of SO₂ - moles of SO₃:
1,34 moles - 0,62 moles =
<em>0,72 moles of SO₂ remain</em>
I hope it helps!