Respuesta :
Answer: Amount of water produced is 8.6778 grams.
Explanation:
To calculate the amount of water produced, we first need to find the limiting reagent and for it we need to find the number of moles. To calculate the moles, we use the following equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ...(1)
- Moles of [tex]C_2H_6[/tex]:
Given mass of [tex]C_2H_6[/tex] = 7 grams
Molar mass of [tex]C_2H_6[/tex] = 30 g/mol
Putting values in above equation, we get:
[tex]\text{Number of moles}=\frac{7g}{30g/mol}=0.2333moles[/tex]
- Moles of [tex]O_2[/tex]:
Given mass of [tex]O_2[/tex] = 18 grams
Molar mass of [tex]O_2[/tex] = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Number of moles}=\frac{18g}{32g/mol}=0.5625moles[/tex]
For the given chemical reaction, the equation follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
By Stoichiometry:
7 moles of oxygen reacts with 2 moles of ethane.
So, 0.0261 moles of lead nitrate are produced by = [tex]\frac{2}{7}\times 0.5625=0.1607moles[/tex] of ethane.
As, the required amount of ethane is less than the given amount. Hence, it is considered as the excess reagent.
Oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
7 moles of oxygen gas produces 6 moles of water.
So, 0.5625 moles of oxygen gas will produce = [tex]\frac{6}{7}\times 0.5625=0.4821moles[/tex] of water.
- Now, to calculate the mass of water produced, we use equation 1.
Molar mass of water = 18 g/mol
Moles of water produced = 0.4821 moles
Putting values in the equation, we get:
[tex]0.4821mol=\frac{\text{Mass of water}}{18g/mol}[/tex]
Mass of water produced = 8.6778 grams.
Hence, amount of water produced is 8.6778 grams.
