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Find the weight of NH3 in grams when 2.8...

Find the weight of `NH_3` in grams when 2.8 kg `N_2` reacts with 1Kg `H_2` ?

A

6000g

B

3400g

C

3000g

D

5666g

Text Solution

AI Generated Solution

The correct Answer is:
To find the weight of ammonia (NH₃) produced when 2.8 kg of nitrogen (N₂) reacts with 1 kg of hydrogen (H₂), we can follow these steps: ### Step 1: Write the balanced chemical equation The balanced reaction for the formation of ammonia from nitrogen and hydrogen is: \[ N_2 + 3H_2 \rightarrow 2NH_3 \] ### Step 2: Calculate the molar masses - Molar mass of \( N_2 \) = 28 g/mol (since nitrogen has an atomic mass of 14 g/mol) - Molar mass of \( H_2 \) = 2 g/mol (since hydrogen has an atomic mass of 1 g/mol) - Molar mass of \( NH_3 \) = 17 g/mol (14 g/mol for nitrogen + 3 g/mol for hydrogen) ### Step 3: Convert the masses from kg to grams - Mass of \( N_2 \) = 2.8 kg = 2800 g - Mass of \( H_2 \) = 1 kg = 1000 g ### Step 4: Determine the number of moles of each reactant - Moles of \( N_2 \): \[ \text{Moles of } N_2 = \frac{2800 \text{ g}}{28 \text{ g/mol}} = 100 \text{ moles} \] - Moles of \( H_2 \): \[ \text{Moles of } H_2 = \frac{1000 \text{ g}}{2 \text{ g/mol}} = 500 \text{ moles} \] ### Step 5: Identify the limiting reagent From the balanced equation, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore, for 100 moles of \( N_2 \), we need: \[ 100 \text{ moles of } N_2 \times 3 \text{ moles of } H_2/\text{mole of } N_2 = 300 \text{ moles of } H_2 \] Since we have 500 moles of \( H_2 \), which is more than enough, \( N_2 \) is the limiting reagent. ### Step 6: Calculate the amount of \( NH_3 \) produced From the balanced equation, 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \). Therefore, 100 moles of \( N_2 \) will produce: \[ 100 \text{ moles of } N_2 \times 2 \text{ moles of } NH_3/\text{mole of } N_2 = 200 \text{ moles of } NH_3 \] ### Step 7: Convert moles of \( NH_3 \) to grams Now, we convert the moles of \( NH_3 \) to grams: \[ \text{Mass of } NH_3 = 200 \text{ moles} \times 17 \text{ g/mol} = 3400 \text{ g} \] ### Final Answer The weight of \( NH_3 \) produced is **3400 grams**. ---
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