liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). I
1 answer:
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:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>