Answer:
The % yield is 27.0 %
Explanation:
<u>Step 1: </u>Data given
Mass of sulfuric acid = 4.9 grams
Mass of sodium hydroxide = 7.8 grams
Mass of sodium sulfate produced = 1.92 grams
Molar mass H2SO4 = 98.08 g/mol
Molar mass NaOH = 40 g/mol
Molar mass Na2SO4 = 142.04 g/mol
<u>Step 2: </u>The balanced equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
<u>Step 3</u>: Calculate moles H2SO4
Moles H2SO4 = Mass H2SO4 / Molar mass H2SO4
Moles H2SO4 = 4.9 grams / 98.08 g/mol =
Moles H2SO4 = 0.05 moles
<u>Step 4:</u> Calculate moles NaOH
Moles NaOH = 7.8 grams / 40 g/mol
Moles NaOH = 0.195 moles
<u>Step 5</u>: Calculate limiting reactant
For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4 and 2 moles H2O
H2SO4 is the limiting reactant. It will completely be consumed (0.05 moles).
NaOH is in excess. There will react 2*0.05 = 0.1 moles
There will remain 0.195 -0.1 = 0.095 moles NaOH
<u>Step 6:</u> Calculate moles Na2SO4
For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4
For 0.05 moles H2SO4, we have 0.05 moles Na2SO4
<u>Step 7:</u> Calculate mass of Na2SO4
Mass Na2SO4 = Moles Na2SO4 * Molar mass Na2SO4
Mass = 0.05 moles * 142.04 g/mol = 7.102
This is the theoretical yield
<u>Step 8:</u> Calculate the percent yield of Na2SO4
% yield = (actual yield / theoretical yield) * 100%
% yield = (1.92 / 7.102) *100% = 27.0 %
The % yield is 27.0 %
