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How many grams of N2 can be produced when 4.40 g of O2 reacts?4NH3(g)+3O2(g)→2N2(g)+6H2O(g)

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Step-by-step explanation:

4 NH₃ (g) + 3 O₂ (g) ----> 2 N₂ (g) + 6 H₂O (g)

We have to determine the mass in grams of nitrogen gas that can be produced from 4.40 g of oxygen gas.

First we have to convert the mass of oxygen gas into moles using the molar mass of it.

molar mass of O = 16.00 g/mol

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of O₂ = 4.40 g * 1 mol/(32.00 g)

moles of O₂ = 0.1375 moles

4 NH₃ (g) + 3 O₂ (g) ----> 2 N₂ (g) + 6 H₂O (g)

According to the coefficients of the equation 3 moles of O₂ will produce 2 moles of N₂. Then the molar ratio between them is 3 to 2. We can use this ratio to find the number of moles of N₂ that will be produced by 0.1375 moles of O₂.

3 moles of O₂ = 2 moles of N₂ molar ratio

moles of N₂ = 0.1375 moles of O₂ * 2 moles of N₂/(3 moles of O₂)

moles of N₂ = 0.0917 moles

And finally we can convert the moles back to grams using the molar mass of nitrogen gas.

molar mass of N = 14.01 g/mol

molar mass of N₂ = 2 * 14.01 g/mol

molar mass of N₂ = 28.02 g/mol

mass of N₂ = 0.0917 moles * 28.02 g/mol

mass of N₂ = 2.57 g

Answer: the mass of nitrogen gas that can be produced is 2.57 g.

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