Answer:
The answer to your question is there will be 20.5 g left of HCl
Explanation:
Data
HCl = 31.4 g
NaOH = 12 g
Excess of HCl = ?
Balanced chemical reaction
HCl + NaOH ⇒ NaCl + H₂O
Molar weight of HCl = 1 + 35.5 = 36.5 g
Molar weight of NaOH = 23 + 16+ 1 = 40 g
Calculate the limiting reactant
theoretical yield HCl/NaOH = 36.5/40 = 0.9125
experimental yield HCl/NaOH = 31.4/12 = 2.62
Conclusion
The limiting reactant is NaOH because the experimental yield increases.
Calculate the mass of Excess reactant
36.5 g of HCl ---------------- 40 g of NaOH
x ---------------- 12 g of NaOH
x = (12 x 36.5) / 40
x = 438 / 40
x = 10.95 g of HCl
Excess HCl = 31.4 - 10.95
= 20.5 g
Answer:
There will remain 20.5 grams of hydrochloric acid
Explanation:
Step 1: Data given
Mass of hydrochloric acid = 31.4 grams
Mass of sodium hydroxide = 12.0 grams
Molar mass hydrochloric acid (HCl) = 36.46 g/mol
Molar mass sodium hydroxide (NaOH) = 40.0 g/mol
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles HCl = 31.4 grams / 36.46 g/mol
Moles HCl = 0.861 moles
Moles NaOH = 12.0 grams / 40.0 g/mol
Moles NaOH = 0.3 moles
Step 4: Calculate the limiting reactant
The limiting reactant is NaOH. It will completely be consumed (0.3 moles). Hcl is in excess. There will react 0.3 moles. There will remain 0.861 - 0.3 = 0.561 moles
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass
Mass HCl = 0.561 moles * 36.46 g/mol
Mass HCl = 20.5 grams
There will remain 20.5 grams of hydrochloric acid
