We will determine the moles and grams of $O_2$ required to react with 58.6 g of $NH_3$ using the given reaction:

$NH_3 + O_2 \rightarrow NO + H_2O$

Step 1: Balance the Chemical Equation

The given reaction is unbalanced, so we balance it first:

$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$

Now, we use this balanced equation to find the required oxygen.

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Step 2: Find the Moles of $NH_3$

The molar mass of ammonia ($NH_3$) is:

$\text{Molar mass of } NH_3 = 14.01 + (3 \times 1.008) = 17.03 \text{ g/mol}$

Moles of $NH_3$ in 58.6 g:

$\text{Moles of } NH_3 = \frac{58.6}{17.03} \approx 3.44 \text{ moles}$

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Step 3: Use Stoichiometry to Find Moles of $O_2$

From the balanced equation:

$4 \text{ moles of } NH_3 \text{ react with } 5 \text{ moles of } O_2$

Using this ratio:

$\text{Moles of } O_2 = 3.44 \times \frac{5}{4} = 4.30 \text{ moles}$

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Step 4: Convert Moles of $O_2$ to Grams

The molar mass of $O_2$ is:

$\text{Molar mass of } O_2 = 2 \times 16.00 = 32.00 \text{ g/mol}$

Mass of $O_2$ needed:

$\text{Mass of } O_2 = 4.30 \times 32.00 = 137.6 \text{ g}$

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Final Answer

• Moles of $O_2$ required: 4.30 moles
• Mass of $O_2$ required: 137.6 g

Would you like a step-by-step breakdown of the calculations?

Related Questions

1. What is the mass of NO produced in this reaction?
2. How many liters of $O_2$ (at STP) are required for this reaction?
3. What is the mass of water formed in this reaction?
4. What would happen if there was only 100 g of $O_2$ available?
5. How much excess reactant would remain if 80 g of $NH_3$ were used?

Tip:

In any stoichiometry problem, always start with a balanced chemical equation to ensure the correct mole ratio.