Respuesta :
the combustion reaction for butane is as follows ;
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
Stoichiometry of butane to oxygen is 2:13
Number of butane moles reacted = 5.23 g / 58 g/mol = 0.090 mol
number of Oxygen moles reacted = 27.1 g/ 32 g/mol = 0.84 mol
we have to find which is the limiting reactant
if butane is the limiting reactant,
if 2 mol of butane reacts with 13 mol of O₂
then 0.09 mol of butane reacts with -13/2 x 0.090 mol = 0.58 mol
there are 0.84 mol of oxygen present but only 0.58 mol reacts therefore butane is the limiting reactant.
stoichiometry of butane to water is 2:10 = 1:5
the number of water moles formed = 0.090 mol x 5 = 0.45 mol
mass of water formed = 0.45 mol x 18 g/mol = 8.1 g
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
Stoichiometry of butane to oxygen is 2:13
Number of butane moles reacted = 5.23 g / 58 g/mol = 0.090 mol
number of Oxygen moles reacted = 27.1 g/ 32 g/mol = 0.84 mol
we have to find which is the limiting reactant
if butane is the limiting reactant,
if 2 mol of butane reacts with 13 mol of O₂
then 0.09 mol of butane reacts with -13/2 x 0.090 mol = 0.58 mol
there are 0.84 mol of oxygen present but only 0.58 mol reacts therefore butane is the limiting reactant.
stoichiometry of butane to water is 2:10 = 1:5
the number of water moles formed = 0.090 mol x 5 = 0.45 mol
mass of water formed = 0.45 mol x 18 g/mol = 8.1 g
Answer: The theoretical yield of water is 1.18 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For butane:
Given mass of butane = 5.23 g
Molar mass of butane = 58.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of butane}=\frac{5.23g}{58.12g/mol}=0.090mol[/tex]
- For oxygen gas:
Given mass of oxygen gas = 27.1 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{2.71g}{32g/mol}=0.085mol[/tex]
The chemical equation for the reaction of butane and oxygen gas follows:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
By Stoichiometry of the reaction:
13 moles of oxygen gas reacts with 2 moles of butane
So, 0.085 moles of oxygen gas will react with = [tex]\frac{2}{13}\times 0.085=0.0131mol[/tex] of butane
As, given amount of butane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
13 moles of oxygen gas produces 10 moles of water
So, 0.085 moles of oxygen gas will produce = [tex]\frac{10}{13}\times 0.085=0.0654moles[/tex] of water
Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 0.0654 moles
Putting values in equation 1, we get:
[tex]0.0654mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0654mol\times 18g/mol)=1.18g[/tex]
Hence, the theoretical yield of water is 1.18 grams
