Astandard 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.

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:

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

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

answer: your mm

explanation:

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explanation: