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) + 2 H2O(ℓ).
How much NH4NO3 is required to produce
52.1 g of N2O?
Answer in units of g.
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To find out how much NH4NO3 is required to produce 52.1 g of N2O, we need to use stoichiometry and the balanced chemical equation for the decomposition of ammonium nitrate.
The balanced chemical equation is: NH4NO3(s) → N2O(g) + 2 H2O(ℓ)
From the equation, we can see that the molar ratio between NH4NO3 and N2O is 1:1. This means that for every 1 mole of NH4NO3, 1 mole of N2O is produced.
To proceed, we need to determine the molar mass of N2O. Nitrogen (N) has a molar mass of 14.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since there are two oxygen atoms in N2O, the molar mass of N2O is:
Molar mass of N2O = (2 * molar mass of N) + molar mass of O
= (2 * 14.01 g/mol) + 16.00 g/mol
= 44.03 g/mol
Now, we can set up a proportion to find the amount of NH4NO3 required:
(52.1 g N2O) * (1 mol NH4NO3/44.03 g N2O) = x mol NH4NO3
Solving for x gives us the number of moles of NH4NO3 required to produce 52.1 g of N2O. We can use the molar mass of NH4NO3 to convert this to grams:
x * (80.05 g NH4NO3/1 mol NH4NO3) = grams of NH4NO3
Finally, we can substitute the calculated value of x into the equation to find the answer in grams:
(52.1 g N2O) * (1 mol NH4NO3/44.03 g N2O) * (80.05 g NH4NO3/1 mol NH4NO3) = grams of NH4NO3
By plugging in the values and performing the calculations, you can find the amount of NH4NO3 required to produce 52.1 g of N2O.