The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>
First, let's look at the equation of the reaction:
The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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Answer:
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula
The outer core of the Earth is a fluid layer about 2,300 km (1,400 mi) thick and composed of mostly iron<span> and</span>nickel<span> that lies above Earth's solid inner core and below its mantle. Its outer boundary lies 2,890 km (1,800 mi) beneath Earth's surface.</span>
The density of hydrogen : ρ = 0.0892 g/L
<h3>Further explanation</h3>
Given
mass of Hydrogen : 0.446 g
Volume = 5 L
Required
The density
Solution
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
The unit of density can be expressed in g/cm³, kg/m³, or g/L
Density formula:
Input the value :
ρ = m : V
ρ = 0.446 g : 5 L
ρ = 0.0892 g/L
Answer:
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