The moles of oxygen required to burn Butane is 6 moles.
<h3>What is a Combustion Reaction?</h3>
A reaction in which fuel gets oxidised by an oxidising agent producing a large amount of heat is called a combustion reaction.
In this question
Butane is burnt with oxygen
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol
Balanced equation for the reaction:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Mole ratio C₄H₁₀ : O₂ = 2 : 13
The given mass = 54grams
moles = 54/58 = 0.93 moles
The mole of oxygen required =
0.93/ x = 2/13
0.93*13/2 = x
x = 6.045 moles
Therefore 6 moles of oxygen are required to burn Butane.
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Explanation:
Let relative ratio of one isotope (62.94 u) be X
Then, relative ratio of other isotope (64.93) will be (1 - X)
Now,
(62.94)x + (64.93)(1 - x) = 63.55
1.99x = 1.38
X = 0.69
<u>Relative Abundance</u> :
(62.94 u) isotope = 69 %
(64.93 u) isotope = 31 %
>4Dt. 1950 y fueron. 45Dys Imox x Na motu cues G.91 %10° mole cultes. 3) How many grams are there in 2.3 x 1024 atoms of silver? 2, 3x 10°tatoms x Imol.<
Answer:
0.35 g Li
Explanation:
H2SO4 + 2Li -> Li2SO4 + H2
7 g Li -> 2 g H2
x -> 0.10 g H2
x= (0.10 g H2 * 7 g Li)/ 2 g H2 x= 0.35 g Li
