Respuesta :
Answer:
1.63 g of H₂O is the theoretical yield
Explanation:
We determine the reactants for the reaction:
HCl, NaOH
We determine the products for the reaction:
H₂O, NaCl
The equation for this neutralization reaction is:
HCl (aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
We use both masses, from both reactants to determine the limiting.
First, we convert the mass to moles.
3.3 g . 1mol / 36.45 g = 0.090 moles of HCl
6.6 g . 1mol / 40 g = 0.165 moles of NaOH
Ratio is 1:1, so for 0.165 moles of hydroxide I need the same amount of acid. I have 0.090 HCl so the acid is the limiting reagent.
Let's work with stoichiometry. Ratio is 1:1, again.
1 mol of acid can produce 1 mol of water
Therefore, 0.090 moles of acid must produce 0.090 moles of H₂O
We convert the moles to mass, to define the theoretical yield
0.090 mol . 18g / 1 mol = 1.63 g
Answer:
The theoretical yield of H2O is 1.63 grams
Explanation:
Step 1: Data given
Mass of hydrochloric acid = 3.3 grams
Mass of sodium hydroxide = 6.6 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 = 3.3 grams / 36.46 g/mol
Moles HCl = 0.0905 moles
Moles NaOH = 6.6 grams / 40.0 g/mol
Moles NaOH = 0.165 moles
Step 4: Calculate the limiting reactant
HCl is the limiting reactant. There will react 0.0905 moles. NaOH is in excess. There will react 0.0905 moles moles. There will remain 0.165 - 0.0905 = 0.0745 moles
Step 5: Calculate moles H2O
For 1 mol HCl we need 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O
For 0.0905 moles HCl we'll have 0.0905 moles H2O
Step 6: Calculate mass H2O
Mass H2O = moles * molar mass
Mass H2O = 0.0905 * 18.02 g/mol
Mass H2O = 1.63 grams
The theoretical yield of H2O is 1.63 grams
