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QuestionCalculate grams of nitrogen gas produced from 30.9 grams of ammonium nitrite using their molar masses.

Studdy Solution

STEP 1

Assumptions1. The given reaction is balanced. . The reaction goes to completion.
3. The molar mass of ammonium nitrite (NH4NO) is64.06 g/mol.
4. The molar mass of nitrogen gas () is28.01 g/mol.
5. The mass of ammonium nitrite given is30.9 g.

STEP 2

First, we need to find the number of moles of ammonium nitrite. We can do this by dividing the mass of ammonium nitrite by its molar mass.
Moles of NH4NO2=Mass of NH4NO2Molar mass of NH4NO2\text{Moles of NH4NO2} = \frac{\text{Mass of NH4NO2}}{\text{Molar mass of NH4NO2}}

STEP 3

Now, plug in the given values for the mass and molar mass of ammonium nitrite to calculate the number of moles.
Moles of NHNO2=30.9 g64.06 g/mol\text{Moles of NHNO2} = \frac{30.9\, g}{64.06\, g/mol}

STEP 4

Calculate the number of moles of ammonium nitrite.
Moles of NH4NO2=30.9 g64.06 g/mol≈0.482 mol\text{Moles of NH4NO2} = \frac{30.9\, g}{64.06\, g/mol} \approx0.482\, mol

STEP 5

From the balanced chemical equation, we can see that one mole of ammonium nitrite produces one mole of nitrogen gas. Therefore, the number of moles of nitrogen gas produced is the same as the number of moles of ammonium nitrite.
Moles of N2=Moles of NH4NO2=0.482 mol\text{Moles of N2} = \text{Moles of NH4NO2} =0.482\, mol

STEP 6

Now that we have the number of moles of nitrogen gas, we can find the mass of nitrogen gas produced. This can be done by multiplying the number of moles of nitrogen gas by its molar mass.
Mass of N2=Moles of N2×Molar mass of N2\text{Mass of N2} = \text{Moles of N2} \times \text{Molar mass of N2}

STEP 7

Plug in the values for the number of moles and the molar mass of nitrogen gas to calculate the mass.
Mass of N2=0.482 mol×28.01 g/mol\text{Mass of N2} =0.482\, mol \times28.01\, g/mol

STEP 8

Calculate the mass of nitrogen gas produced.
Mass of N2=0.482 mol×28.01 g/mol≈13.5 g\text{Mass of N2} =0.482\, mol \times28.01\, g/mol \approx13.5\, gUpon the complete reaction of30. grams of ammonium nitrite, approximately13.5 grams of nitrogen gas will be formed.

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