A 3 sig figs
B 6 sig figs
C 2 sig figs
D 5 sig figs
C, A, D, B
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%
Solids— have tightly packed molecules, structured shape
Liquid— have no definite shape and the molecules are not tightly packed like a solid, they are a lot more loose.
Gas— No structure at all, molecules are free flowing
Sorry I couldn’t be a better help.. but I hope this gives you a bit of an idea :)
Answer:
Molarity = 0.154 M
Explanation:
Given data:
Volume of solution = 8.42 × 10² mL ( 8.42 × 10² /1000 = 0.842 L)
Mass of potassium iodide = 22.4 g
Molarity of solution = ?
Solution:
Number of moles of potassium iodide:
Number of moles = mass/molar mass
Number of moles = 22.4 g/ 166.0 g/mol
Number of moles = 0.13 mol
Molarity:
Molarity = number of moles / volume in L
Molarity = 0.13 mol / 0.842 L
Molarity = 0.154 M
