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A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, Na2CO3. The balanced equation is:
2 HNO3 (aq) + Na2CO3 (s) ➝ 2 NaNO3 (aq) + H2O (l) + CO2 (g)
Referring to Example Exercise 2, provide in three significant figures:
a) # moles of Na2CO3
b) # moles of HNO3
c) molarity of HNO3 if 33.25 mL are required to reach a permanent endpoint.

Respuesta :

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Answer:

a) 4.01x10⁻³ mol Na₂CO₃

b) 8.02x10⁻³ mol HNO₃

c) 0.241 M

Explanation:

2HNO₃ (aq) + Na₂CO₃ (s) ➝ 2NaNO₃ (aq) + H₂O (l) + CO₂ (g)

a) To calculate the moles of Na₂CO₃ we use its molecular weight:

  • MW Na₂CO₃ = 23*2 + 12 * 16 * 3 = 106 g/mol
  • 0.425 gNa₂CO₃ ÷ 106 g/mol = 4.01x10⁻³ mol Na₂CO₃

b) To calculate the moles of HNO₃ we use the stoichiometric ratio between Na₂CO₃ and HNO₃:

  • 4.01x10⁻³ mol Na₂CO₃ * [tex]\frac{2molHNO_{3}}{1molNa_{2}CO_{3}}[/tex] = 8.02x10⁻³ mol HNO₃

c) We divide the moles of HNO₃ by the volume in order to calculate molarity:

  • 33.25 mL ⇒ 33.25/1000 = 0.03325 L
  • 8.02x10⁻³ mol HNO₃ / 0.03325 L = 0.241 M

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