Respuesta :
Answer : The maximum mass of [tex]H_2O[/tex] produced will be, 3.24 grams
Explanation : Given,
Mass of [tex]CH_4[/tex] = 1.44 g
Mass of [tex]O_2[/tex] = 9.5 g
Molar mass of [tex]CH_4[/tex] = 16 g/mole
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]CH_4[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{1.44g}{16g/mole}=0.09moles[/tex]
[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{9.5g}{32g/mole}=0.29moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that
As, 1 moles of [tex]CH_4[/tex] react with 2 mole of [tex]O_2[/tex]
So, 0.09 moles of [tex]CH_4[/tex] react with [tex]2\times 0.09=0.18[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CH_4[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_2O[/tex].
As, 1 mole of [tex]CH_4[/tex] react to give 2 moles of [tex]H_2O[/tex]
So, 0.09 moles of [tex]CH_4[/tex] react to give [tex]2\times 0.09=0.18[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex].
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(0.18mole)\times (18g/mole)=3.24g[/tex]
Therefore, the maximum mass of [tex]H_2O[/tex] produced will be, 3.24 grams.
The maximum mass of water (H₂O) produced from the reaction between 1.44 g of CH₄ and 9.5 g of O₂ is 3.24 g
We'll begin by obtaining the limiting reactant of the reaction. This is illustrated:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Molar mass of CH₄ = 12 + (1×4) = 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
Molar mass of H₂O = (1×2) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 2 × 18 = 36 g
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Therefore,
1.44 g of CH₄ will react with = [tex]\frac{1.44 * 64}{16}\\\\[/tex] = 5.76 g of O₂.
From the above calculation, only 5.76 g of O₂ out of 9.5 g reacted completely with 1.44 g of CH₄.
Thus, CH₄ is the limiting reactant and O₂ is the excess reactant.
Finally, we shall determine the maximum mass of water (H₂O) produced from the reaction. This can be obtained as follow:
From the balanced equation above,
16 g of CH₄ reacted to produce 36 g of H₂O.
Therefore,
1.44 g of CH₄ will react to produce = [tex]\frac{1.44 * 36}{16}\\\\[/tex] = 3.24 g of H₂O.
Thus, the maximum mass of water (H₂O) produced from the reaction is 3.24 g
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