Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate is decomposed

according to the reaction
NH4NO3(s) −! N2O(g) + 2H2O(§¤) .
How much NH4NO3 is required to produce
54.5 g of N2O?
Answer in units of g.

To determine the amount of NH4NO3 required to produce 54.5 g of N2O, we need to calculate the molar mass of NH4NO3 and then use stoichiometry to find the corresponding mass.

The molar mass of NH4NO3 can be calculated by adding up the atomic masses of its constituent elements:

Molar mass of NH4NO3 = (1 * molar mass of N) + (4 * molar mass of H) + (1 * molar mass of NO3)

Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of NO3 = (1 * molar mass of N) + (3 * molar mass of O)

Molar mass of O = 16.00 g/mol

Now, let's calculate the molar mass of NO3 and then use it to find the molar mass of NH4NO3:

Molar mass of NO3 = (1 * molar mass of N) + (3 * molar mass of O)
= (1 * 14.01 g/mol) + (3 * 16.00 g/mol)
= 62.01 g/mol

Molar mass of NH4NO3 = (1 * molar mass of N) + (4 * molar mass of H) + (1 * molar mass of NO3)
= (1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (1 * 62.01 g/mol)
= 80.03 g/mol

Now we can use stoichiometry to find the mass of NH4NO3:

1 mol of NH4NO3 produces 1 mol of N2O

Given:
Mass of N2O = 54.5 g
Molar mass of N2O = 44.01 g/mol

Using the molar mass of N2O, we can calculate the number of moles of N2O:

Number of moles of N2O = Mass of N2O / Molar mass of N2O
= 54.5 g / 44.01 g/mol
= 1.238 mol

Since 1 mol of NH4NO3 produces 1 mol of N2O, the mass of NH4NO3 required can be determined by multiplying the number of moles of N2O by the molar mass of NH4NO3:

Mass of NH4NO3 = Number of moles of N2O * Molar mass of NH4NO3
= 1.238 mol * 80.03 g/mol
= 99.0 g

Therefore, 99.0 g of NH4NO3 is required to produce 54.5 g of N2O.

To determine the amount of NH4NO3 required to produce 54.5 g of N2O, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.

The balanced equation shows that for every 1 mole of NH4NO3, we produce 1 mole of N2O. To calculate the amount of NH4NO3, we need to convert the grams of N2O into moles, and then use the mole ratio from the balanced equation.

First, let's calculate the number of moles of N2O:

Molar mass of N2O = 28.02 g/mol
Number of moles of N2O = Mass of N2O / Molar mass of N2O = 54.5 g / 28.02 g/mol ≈ 1.9457 mol

According to the balanced equation, 1 mole of NH4NO3 produces 1 mole of N2O. Therefore, the number of moles of NH4NO3 required to produce 1.9457 mol of N2O is also 1.9457 mol.

Next, we need to calculate the mass of NH4NO3:

Molar mass of NH4NO3 = (1 × molar mass of N) + (4 × molar mass of H) + (1 × molar mass of N) + (3 × molar mass of O)
≈ (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + (1 × 14.01 g/mol) + (3 × 16.00 g/mol)
≈ 80.04 g/mol

Mass of NH4NO3 = Number of moles of NH4NO3 × Molar mass of NH4NO3
= 1.9457 mol × 80.04 g/mol ≈ 155.70 g

Therefore, approximately 155.70 grams of NH4NO3 are required to produce 54.5 g of N2O.

This looks like another regular stoichometry problem. If you don't know how to do these (I noted this at your posts below) what is it you don't understand?

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