Answer:
- 5.15×10²⁴ molecules of sulfur dioxide
- 3.63×10²³ molecules of carbon monoxide
- 6.02×10²³ molecules of ammonia
Explanation:
We begin from the relation that 1 mol of molecules contains NA of molecules
NA = 6.02×10²³
Now, we make rules of three:
1 mol has 6.02×10²³ molecules, therefore:
8.55 moles of SO₂ must have (8.55 . NA) / 1 = 5.15×10²⁴ molecules of dioxide
0.603 moles of CO must have (0.603 . NA) / 1 = 3.63×10²³ molecules of monoxide
Avogadro's Number of molecules of NH₃ are 6.02×10²³ molecules of ammonia
There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
Answer:
19.4 g of alum, will be its theoretical yield
Explanation:
The reaction is:
2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O
Let's determine the amount of acid.
M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution
M = mmol /mL
M . mL = mmol
We replace: 8.3 mL . 9.9 M = 82.17 mmoles
We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles
Ratio is 4:2
4 moles of sulfuric acid can make 2 moles of alum
By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.
We convert moles to mass:
Molar mass of alum is: 473.52 g/mol.
0.041085 moles . 473.52 g/mol = 19.4 g
