Answer:
1. 0.421 g (C₄H₁₀), 1.51 g (O₂), 1.28 g (CO₂), 0.653 g (H₂O)
2. 4.92 g (C₄H₁₀), 17.6 g (O₂), 14.9 g (CO₂), 7.63 g (H₂O)
3. 6.63 g (C₄H₁₀), 23.7 g (O₂), 20.12 g (CO₂), 10.3 g (H₂O)
4. 7.12 g (C₄H₁₀), 12.1 g (O₂), 10.2 g (CO₂), 8.84 g (H₂O)
5. 252 mg (C₄H₁₀), 903 mg (O₂), 763 mg (CO₂), 390 mg (H₂O)
6. 65 mg (C₄H₁₀), 234 mg (O₂), 198 mg (CO₂), 101 mg (H₂O)
Explanation:
1. First of all, we determine the moles of each reactant.
For the first case:
1.51 g . 1 mol/32 g = 0.0472 moles
Ratio is 13:2, 13 moles of oxygen needs 2 moles of C₄H₁₀ for the combustion,
Therefore 0.0472 mol will react with (0.0472 . 2)/13 = 7.26×10⁻³ mol.
Now we convert the moles to mass:
7.26×10⁻³ mol . 58 g/ 1mol = 0.421 g
Now we use stoichiometry to find the mass of the products.
Ratio is 13:8:10.
13 moles of oxygen can produce 8 moles of CO₂ and 10 moles of water
Then, 0.0472 mol would produce:
(0.0472 . 8)/13 = 0.0290 mol
We convert the moles to mass → 0.0290 mol . 44g /mol = 1.28 g
(0.0472 . 10)/13 = 0.0363 mol
We convert the moles to mass → 0.0363 mol . 18 g /1mol = 0.653 g
2. 4.92 g / 58 g/mol = 0.0848 moles of C₄H₁₀
2 moles of C₄H₁₀ react with 13 moles of O₂
So, 0.0848 moles will react with (0.0848 . 13) / 2 = 0.551 moles
We convert to mass: 0.551 mol . 32 g /mol = 17.6 g
Now we use stoichiometry to find the mass of the products.
Ratio is 13:8:10.
0.551 moles of O₂ will produce:
(0.551 . 8)/13 = 0.339 mol of CO₂
We convert to mass: 0.339 mol . 44g / mol = 14.9 g
(0.551 . 10)/13 = 0.424 mol of H₂O
0.424 mol . 18 g /mol = 7.63 g
3. In this case, we have the mass of one of the product
20.12 g . 1mol / 44 g = 0.457 moles of CO₂
According to stoichiometry:
8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀
Then, 0.457 moles of CO₂ would be produced by:
(0.457 . 13)/ 8 = 0.743 moles of O₂
We convert to mass: 0.743 mol . 32 g/1mol = 23.7 g
(0.457 . 2)/8 = 0.114 moles of C₄H₁₀
We convert to mass: 0.114 mol . 58g/mol = 6.63g
Now we can determine, the mass of produced water:
(0.743 . 10)/13 = 0.571 mol of H₂O . 18g /mol = 10.3 g
4. We convert the moles of water:
8.84 g / 18g/mol = 0.491 moles
According to stoichiometry: 10 moles of water are produced by 13 moles of O₂ and 2 moles of C₄H₁₀
Then 0.491 moles will be produced by:
(0.491 . 10)/ 13 = 0.378 moles of O₂
We convert to mass: 0.378 mol . 32 g/1mol = 12.1 g
(0.491 . 2)/8 = 0.123 moles of C₄H₁₀
We convert to mass: 0.123 mol . 58g/mol = 7.12g
Now we can determine, the mass of produced carbon dioxide:
(0.378 . 8)/13 = 0.232 mol of CO₂ . 44g /mol = 10.2 g
5. Mass of mg, must be converted to grams
252 mg . 1 g/1000 mg = 0.252 g
It is the same as 2.
0.252 g of C₄H₁₀ . 1mol/58 g = 4.34×10⁻³ mol
2 mol of C₄H₁₀ react to 13 moles of O₂ then,
4.34×10⁻³ mol will react with (4.34×10⁻³ mol . 13) / 2 = 0.0282 mol
We convert the grams → 0.0282 mol . 32 g/mol = 0.903 g (903 mg)
0.0282 mol of oxygen will produced:
(0.0282 . 8)/13 = 0.0173 mol of CO₂
We convert to mass: 0.0173 mol . 44g / mol = 0.763 g (763 mg)
(0.0282 . 10)/13 = 0.0217 mol of H₂O
0.0217 mol . 18 g /mol = 0.390 g (390 mg)
6. We define the mass of CO₂ → 198 mg . 1g/1000 mg = 0.198 g
0.198 g / 44g/mol = 4.5×10⁻³ moles of CO₂
According to stoichiometry:
8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀
Then, 4.5×10⁻³ moles of CO₂ would be produced by:
(4.5×10⁻³ . 13)/ 8 = 7.31×10⁻³ moles of O₂
We convert to mass: 7.31×10⁻³ . 32 g/1mol = 0.234 g (234 mg)
(4.5×10⁻³ . 2)/8 = 1.125×10⁻³ moles of C₄H₁₀
We convert to mass: 1.125×10⁻³ mol . 58g/mol = 0.065 g (65 mg)
Now we can determine, the mass of produced water:
(7.31×10⁻³ . 10)/13 = 5.62×10⁻³ mol of H₂O . 18g /mol = 0.101 g (101 mg)
