Answer :
Answer:
To determine the grams of NO formed, we need to use stoichiometry to find the balanced equation's conversion factors.
Let's start by calculating the number of moles of NH₃ and O₂:
Molar mass of NH₃ (ammonia) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
Moles of NH₃ = 25.0 g / 17.03 g/mol = 1.47 mol
Molar mass of O₂ (oxygen) = 2(16.00 g/mol) = 32.00 g/mol
Moles of O₂ = 39.4 g / 32.00 g/mol = 1.23 mol
According to the balanced equation, the mole ratio between NH₃ and NO is 4:4. Therefore, 1.47 moles of NH₃ will produce 1.47 moles of NO.
Similarly, the mole ratio between O₂ and NO is 5:4. So, 1.23 moles of O₂ will produce (4/5) * 1.23 = 0.98 moles of NO.
Now, we need to find the mass of NO formed. To do this, we'll use the molar mass of NO.
Molar mass of NO (nitric oxide) = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
Mass of NO formed = moles of NO * molar mass of NO
= 0.98 mol * 30.01 g/mol
≈ 29.41 g
Therefore, approximately 29.41 grams of NO will be formed.
Final answer:
The reaction between NH3 and O2 will produce approximately 29.52 g of NO. This is calculated based on the stoichiometric proportions of the reactants and the molar masses of the substances involved.
Explanation:
First, we need to calculate the number of moles for each reactant - NH3 and O2. The molar mass of NH3 is around 17 g/mol and the molar mass of O2 is around 32 g/mol. Therefore, the number of moles for NH3 is 25.0 g / 17 g/mol = 1.47 mol and for O2 is 39.4 g / 32 g/mol = 1.23 mol.
We can see in the reaction that the stoichiometric ratio between NH3, O2, and NO is 4:5:4. This means that 1 mol of O2 reacts with 4/5 mol of NH3 to produce 4/5 mol of NO. The reaction appears to be limited by O2 as we have less of it in moles. Hence, the quantity of NO produced is restricted by the quantity of O2 present.
By applying the stoichiometric ratio, the amount of NO formed would be 1.23 mol * 4/5 = 0.984 mol. As the molar mass of NO is 30 g/mol, the number of grams is 0.984 mol * 30 g/mol = 29.52 g. Therefore, approximately 29.52 g of NO will be produced in this reaction given the initial quantities of NH3 and O2.
Learn more about Stoichiometry here:
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