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Sagot :
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
Further explanation
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g
0.137moles of water is formed, 0.0274 moles or 1.5892 g of butane burned and 0.1781 moles or 5.6992 g of O₂ is used.
What is combustion reaction?
Those reaction in which fuel is oxidized by the oxygen molecules and produce carbon dioxide and water molecule.
Given chemical reaction is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Given mass of water = 2.46 grams
Moles will be calculated as:
n = W/M, where
W = given mass
M = molar mass
(a) Moles of water formed is calculated as:
Moles of water n = 2.46g / 18 g/mol = 0.137moles
(b) From the stoichiometry of the reaction, it is clear that:
10 moles of water = produced by 2 moles of butane
0.137 moles of water = produced by 2/10×0.137=0.0274 moles of butane
(c) Weight of butane is calculated by using moles:
W = 0.0274 × 58 g/mol = 1.5892 g
(d) From the stoichiometry of the reaction, it is clear that:
2 moles of butane = react with 13 moles of O₂
0.027 moles of butane = react with 13/2×0.027=0.1781 moles of O₂
(e) Mass of oxygen is calculated as:
W = 0.1781 x 32 g/mol = 5.6992 g
Hence, (a) 0.137moles, (b) 0.0274 moles, (c) 1.5892 g, (d) 0.1781 moles and (e) 5.6992 g.
To know more about combustion reaction, visit the below link:
https://brainly.com/question/9425444
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