Phthalonitrile $\left( C _8 H _4 N_2\right)$ is produced by the ammoxidation of o-xylene ( $C _8 H _{10}$ ) according to the following reaction:
$C_8 H_{10}(l)+O_2(g)+NH_3(g) \rightarrow C_8 H_4 N_2(s)+H_2 O(l)$
How many grams of water would be produced by the complete ammoxidation of 3.98 moles of oxylene?
Answer (2)
The complete ammoxidation of 3.98 moles of o-xylene produces approximately 215.10 grams of water. This is calculated using the balanced chemical equation and stoichiometric ratios. Each mole of o-xylene yields three moles of water, leading to the final mass calculation.
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Balance the chemical equation: C 8 H 10 ( l ) + O 2 ( g ) + N H 3 ( g ) → C 8 H 4 N 2 ( s ) + 3 H 2 O ( l ) .
Determine the stoichiometric ratio between o-xylene and water, which is 1:3.
Calculate the moles of water produced: 3 × 3.98 = 11.94 moles.
Calculate the mass of water produced: 11.94 × 18.015 ≈ 215.10 grams.
Explanation
Analyze the problem and balance the equation The problem states that phthalonitrile ( C 8 H 4 N 2 ) is produced by the ammoxidation of o-xylene ( C 8 H 10 ) according to the reaction: C 8 H 10 ( l ) + O 2 ( g ) + N H 3 ( g ) → C 8 H 4 N 2 ( s ) + H 2 O ( l ) .
We are given that 3.98 moles of o-xylene are completely reacted, and we need to find the mass of water produced. First, we need to balance the chemical equation.
Balance the chemical equation To balance the equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation. The unbalanced equation is: C 8 H 10 ( l ) + O 2 ( g ) + N H 3 ( g ) → C 8 H 4 N 2 ( s ) + H 2 O ( l ) .
Balancing the hydrogen atoms, we have 10 hydrogen atoms on the left and 2 on the right. To balance this, we can put a coefficient of 3 in front of H 2 O :
C 8 H 10 ( l ) + O 2 ( g ) + N H 3 ( g ) → C 8 H 4 N 2 ( s ) + 3 H 2 O ( l ) .
Now we have 10 hydrogen atoms on the left and 4 + 6 = 10 hydrogen atoms on the right. The equation is now balanced.
Determine the stoichiometric ratio From the balanced chemical equation: C 8 H 10 ( l ) + O 2 ( g ) + N H 3 ( g ) → C 8 H 4 N 2 ( s ) + 3 H 2 O ( l ) ,
we can see that 1 mole of o-xylene ( C 8 H 10 ) produces 3 moles of water ( H 2 O ). Therefore, the stoichiometric ratio between o-xylene and water is 1:3.
Calculate moles of water produced Since 3.98 moles of o-xylene are completely reacted, the number of moles of water produced can be calculated using the stoichiometric ratio: Moles of H 2 O = 3 × Moles of C 8 H 10 = 3 × 3.98 = 11.94 moles.
Calculate the molar mass of water To find the mass of water produced, we need to calculate the molar mass of water ( H 2 O ). The molar mass of hydrogen (H) is approximately 1.008 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol. Therefore, the molar mass of water is: Molar mass of H 2 O = 2 × 1.008 + 16.00 = 18.016 g/mol. For simplicity, we can approximate the molar mass of water as 18.015 g/mol.
Calculate the mass of water produced Now we can calculate the mass of water produced by multiplying the number of moles of water by its molar mass: Mass of H 2 O = Moles of H 2 O × Molar mass of H 2 O = 11.94 moles × 18.015 g/mol = 215.0981 g. Rounding to two decimal places, the mass of water produced is approximately 215.10 g.
Final Answer Therefore, the mass of water produced by the complete ammoxidation of 3.98 moles of o-xylene is approximately 215.10 grams.
Examples
In chemical manufacturing, calculating the mass of byproducts like water is crucial for process optimization and waste management. For instance, if a plant produces phthalonitrile from o-xylene, knowing the exact amount of water generated helps in designing efficient separation and disposal systems. This ensures compliance with environmental regulations and improves the overall sustainability of the chemical process. By accurately predicting byproduct quantities, chemical engineers can fine-tune reaction conditions to maximize yield and minimize waste, leading to more cost-effective and environmentally friendly operations. The balanced chemical equation and stoichiometric calculations are fundamental in achieving these goals.
