I’m sorry if I wasted your time but I think it’s alkali metals but I’m
Not sure
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: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.

Number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
To balance this equation, multiply
by 2 on reactant side. Also, multiply
by 2 on product side. Hence, the equation can be rewritten as follows.

Now, the number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
Answer:could it be kilo
Explanation:it looks like it has the least letters