Answer:
- 143.6 g of N₂ ; 3.08×10²⁴ molecules of N₂
- 451.4 g of CO₂ ; 6.17×10²⁴ molecules of CO₂
- 153.9 g of H₂O ; 5.14×10²⁴ molecules of H₂O
- 27.3 g of O₂ ; 5.14×10²³ molecules of O₂
Step-by-step explanation:
This is a reaction of decomposition:
4C₃H₅N₃O₉(s) → 6N₂(g) + 12CO₂(g) + 10H₂O(g) + O₂(g)
4 moles of solid nitroglycerine decompose to 6 moles of nitrogen, 12 moles of carbon dioxide, 10 moles of water vapor and 1 mol of oxygen
We convert mass to moles → 777 g. 1mol/ 227g = 3.42 moles
4 moles of C₃H₅N₃O₉ can decompose to:
6 moles of N₂ ____ 12 moles of CO₂ ___ 10 moles of H₂O ___ 1 mol of O₂
Then, 3.42 moles of C₃H₅N₃O₉ may decompose to:
(3.42 . 6) / 4 = 5.13 moles of N₂
(3.42 . 12) / 4 = 10.26 moles of CO₂
(3.42 . 10) / 4 = 8.55 moles of water vapor
(3.42 . 1) / 4 = 0.855 moles of oxygen
We convert the moles to mass:
5.13 mol . 28 g/mol = 143.6 g of N₂
10.26 mol . 44 g/mol = 451.4 g of CO₂
8.55 mol . 18 g/mol = 153.9 g of H₂O
0.855 mol . 28 g/mol = 27.3 g of O₂
We count the atoms:
5.13 mol . 6.02×10²³ molecules /mol = 3.08×10²⁴ molecules of N₂
10.26 mol . 6.02×10²³ molecules /mol = 6.17×10²⁴ molecules of CO₂
8.55 mol . 6.02×10²³ molecules /mol = 5.14×10²⁴ molecules of H₂O
0.855 mol . 6.02×10²³ molecules /mol = 5.14×10²³ molecules of O₂