How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?
1 answer:
Answer:
n NaHCO3 = 9.6 E-3 mol
Explanation:
balanced reaction:
- 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
- assuming a concentration of H2SO4 6M....normally worked in the lab
⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4
according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)
⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )
⇒ ,mol NaHCO3 = 9.6 E-3 mol
So 9.6 E-3 mol NaHCO3, are the minimun moles necessary to neutralize the acid.
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