Question

PIU

1. Laughing gas in
produced wh
shing gas (nitrous oxide, N,O) is sometimes used as an anesthetic in dentistry. It is
duced when ammonium nitrate is decomposed according to the following reaction.
NH,NO3(s) N2O(g) + 2H,O(1)
How many grams of NH NO, are required to produce 33.0 g N,O?
How many grams of water are produced in this reaction?

Answer 1

alot

Step-by-step explanation:

Answer 2

`\large \boxed{\text{60.0 g NH$_(4)$NO$_(3)$; 27.0 g of H$_(2)$O}}`

Step-by-step explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

MM: 80.04 44.01 18.02

NH₄NO₃ ⟶ N₂O + 2H₂O

m/g: 33.0

1. Mass of NH₄NO₃

(a) Moles of N₂O

`\text{Moles of N$_(2)$O} = \text{33.0 g N$_(2)$O}* \frac{\text{1 mol N$_(2)$O}}{\text{44.01 g N$_(2)$O}}= \text{0.7498 mol N$_(2)$O}`

(b) Moles of NH₄NO₃

`\text{Moles of NH$_(4)$NO$_(3)$} = \text{0.7498 mol N$_(2)$O} * \frac{\text{1 mol NH$_(4)$NO$_(3)$}}{\text{1 mol N$_(2)$O}} = \text{0.7498 mol NH$_(4)$NO$_(3)$}`

(c) Mass of NH₄NO₃

`\text{Mass of NH$_(4)$NO$_(3)$} =\text{0.7498 mol NH$_(4)$NO$_(3)$} * \frac{\text{80.04 g NH$_(4)$NO$_(3)$}}{\text{1 mol NH$_(4)$NO$_(3)$}} = \textbf{60.0 g NH$_(4)$NO$_(3)$}`

2. Mass of H₂O

(a) Moles of H₂O

`\text{Moles of H$_(2)$O} = \text{0.7498 mol N$_(2)$O} * \frac{\text{2 mol H$_(2)$O}}{\text{1 mol N$_(2)$O}} = \text{1.500 mol H$_(2)$O}`

(c) Mass of H₂O

`\text{Mass of H$_(2)$O} =\text{1.500 mol H$_(2)$O} * \frac{\text{18.02 g H$_(2)$O}}{\text{1 mol H$_(2)$O}} = \textbf{27.0 g H$_(2)$O}\\\\\text{The reaction requires = $\large \boxed{\textbf{60.0 g NH$_(4)$NO$_(3)$}}$ and produces $\large \boxed{\textbf{27.0 g of H$_(2)$O}}$}`