Chemistry HELP!?
2016-05-11 4:35 am
If the molarity of the acetic acid you had access to was 3.0M. How many milimeters of acetic acid would you need to fully react 5.00g sodium bicarbonate?
回答 (1)
2016-05-11 5:06 am
Molar mass of NaHCO₃ = 22.99 + 1.01 + 12.01 + 16.00×3 = 84.01 g/mol
NaHCO₃ + HCl → NaCl + H₂O + CO₂
Mole ratio NaHCO₃ : HCl = 1 : 1
No. of moles of NaHCO₃ = (5.00 g) / (84.01 g/mol) = 0.05952 mol
No. of moles of HCl = 0.05952 mol
Volume of HCl = (0.05952 mol) / (3.0/1000 mol/mL) = 19.84 mL
NaHCO₃ + HCl → NaCl + H₂O + CO₂
Mole ratio NaHCO₃ : HCl = 1 : 1
No. of moles of NaHCO₃ = (5.00 g) / (84.01 g/mol) = 0.05952 mol
No. of moles of HCl = 0.05952 mol
Volume of HCl = (0.05952 mol) / (3.0/1000 mol/mL) = 19.84 mL
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