Respuesta :
Answer:
5.28g
Explanation:
The equation for the reaction between is given below:
HCl + NaOH —> NaCl + H2O
Let us determine which will be the limiting reactant.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl = 3.3g
Number of mole = Mass /Molar Mass
Number of mole of HCl = 3.3/36.5 = 0.09mole
Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH = 4.6g
Number of mole = Mass /Molar Mass
Number of mole of NaOH = 4.6/40 = 0.115mole
From the equation,
1mole of HCl required 1mole of NaOH.
Therefore, 0.09mole of HCl will also require 0.09mole of NaOH.
Now, we see that NaOH is excess and HCl is limiting.
Now we can calculate the theoretical yield as follows:
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
From the equation,
36.5g of HCl produced 58.5g of NaCl.
Therefore, 3.3g of HCl will produce = (3.3 x 58.5)/36.5 = 5.28g
Therefore, the theoretical yield of NaCl is 5.28g
Answer:
The theoretical yield of sodium chloride is 5.3 grams
Explanation:
Step 1: Data given
Mass of hydrochloric acid (HCl) = 3.3 grams
Mass of sodium hydroxide (NaOH) = 4.6 grams
Molar mass HCl = 36.46 g/mol
Molar mass 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 = 4.6 grams / 40.0 g/mol
Moles NaOH = 0.115 moles
Step 4: Calculate limiting reactant
For 1 mol HCl we need 1 mol NaOH to produce1 mol NaCl and 1 mol H2O
HCl is the limiting reactant.It will completely be consumed (0.0905 moles). NaOh is in excess. There will react 0.0905 moles. There will remain 0.115 - 0.0905 = 0.0245 moles NaOH
Step 5: Calculate moles NaCl
For 1 mol HCl we need 1 mol NaOH to produce1 mol NaCl and 1 mol H2O
For0.0905 moles HCl we'll have 0.0905 moles NaCl
Step 6: Calculate the theoretical yield of NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.0905 * 58.44 g/mol
Mass NaCl = 5.3 grams
The theoretical yield of sodium chloride is 5.3 grams
