Answer:
Percent yield of reaction is<em> 150%.</em>
Explanation:
Given data:
Percent yield = ?
Actual yield of SO₃ = 586.0 g
Mass of SO₂ = 705.0 g
Mass of O₂ = 80.0 g
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₂:
Number of moles = mass/ molar mass
Number of moles = 586.0 g/ 64.1 g/mol
Number of moles = 9.1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 80.0 g/ 32g/mol
Number of moles = 2.5 mol
Now we will compare the mole of SO₃ with O₂ and SO₂.
SO₂ : SO₃
2 : 2
9.1 : 9.1
O₂ : SO₃
1 : 2
2.5 : 2×2.5 = 5
The number of moles of SO₃ produced by oxygen are less it will limiting reactant.
Theoretical yield of SO₃:
Mass = number of moles × molar mass
Mass = 5 mol × 80.1 g/mol
Mass = 400.5 g
Percent yield of reaction:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 586.0 g/ 400.5 g× 100
Percent yield = 1.5× 100
Percent yield = 150%
Answer:
9.64g
Explanation:
The balanced equation for the reaction is given below:
2NH3 (g) —> 3H2 (g) + N2 (g)
Next, we need to calculate the mass NH3 that decomposed and the mass of H2 produced from the balanced equation. This is illustrated below:
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 that decomposed from the balanced equation = 2 x 17 = 34g
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 produced from the balanced equation = 3 x 2 = 6g.
Now, we can obtain the mass of H2 formed from 54.6g of NH3 as follow:
From the balanced equation above,
34g of NH3 decomposed to produce 6g of H2.
Therefore, 54.6g of NH3 will decompose to produce = (54.6x6)/34 = 9.64g of H2
Therefore, 9.64g of H2 can be obtained from 54.6g of NH3.
Answer:
it is b just trust me man I know
