Answer:
it is actually b because i did this i picked b and got it right
Explanation:
To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%
<span>Grams of solute per 100 grams of water</span>
