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Margaret [11]
3 years ago
7

Nitrous oxide (n2o), or laughing gas, is commonly used as an anesthetic in dentistry and surgery. how many moles are present in

8.4 liters of nitrous oxide at stp?
Chemistry
1 answer:
Gelneren [198K]3 years ago
5 0

Answer is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.

V(N₂O) = 8.4 L.

V(N₂O) = n(N₂O) · Vm.

Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.

n(N₂O) = 8.4 L ÷ 22.4 L/mol.

n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>

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Answer:

\large \boxed{\text{0.1767 mol/L}}$

Explanation:

(a) Balanced equation

3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O  

(b) Moles of NaOH

\text{Moles of  NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}

(c) Moles of H₃PO₄

The molar ratio is 1 mol H₃PO₄:3 mol NaOH.

\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}

(d) Molar concentration of H₃PO₄

c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}

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