Question

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g) Complete the following table

Answer 1

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)

Step-by-step 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)