Question

How many grams of NO are required to produce 145 g of N2 in the following reaction?
4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)
a. 155 g
b. 186 g
c. 125 g
d. 145 g
e. 129 g

Answer 1

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

$145g \times \frac{1mol}{28.01 g} =5.18 mol$

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

$5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO$

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

$6.22mol \times \frac{30.01g}{mol} =186 g$

Answer 2

The mass of NO required to produce 145 g of N₂ is 186 g.

The correct answer to the question is Option B. 186 g

We'll begin by calculating the mass of NO that reacted and the mass of N₂ produced from the balanced equation.

4NH₃ + 6NO —> 5N₂ + 6H₂O

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 6 × 30 = 180 g

Molar mass of N₂ = 14 × 2 = 28 g/mol

Mass of N₂ from the balanced equation = 5 × 28 = 140 g

From the balanced equation above,

140 g of N₂ were produce by 180 g of NO

Finally, we shall determine the mass of NO required to produce 145 g of N₂

From the balanced equation above,

140 g of N₂ were produce by 180 g of NO

Therefore,

145 g of N₂ will be produce by = (145 × 180) / 140 = 186 g of NO

Thus, 186 g of NO is required for the reaction.

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