What mass of potassium sulphate is produced in this reaction
Answers
Answer:
How do you calculate the mass of potassium sulphate produced when 2.0 g of potassium carbonate reacts with an excess of diluted sulphuric acid?
Edit: many thanks to the comments which point out I missed multiplying the 39.1 by 2.
First write the chemical equation.
K2CO3 + H2SO4 -> K2SO4 + CO2 + H2O
Number of moles of K2CO3
=mass of K2CO3/molar mass of K2CO3
=2.0/(39.1 x 2+12.0+16.0 x 3)
=0.0145 mol
K2CO3 is the limiting reactant here.
Mole ratio of K2SO4 to K2CO3 = 1:1
Hence,
Number of moles of K2SO4
=0.0145 mol
Mass of K2SO4
=0.0145 x molar mass of K2SO4
=0.0145 x (39.1 x 2+32.1+16.0 x 4)
=2.52 g
Answer:
First write the chemical equation.
K2CO3 + H2SO4 -> K2SO4 + CO2 + H2O
Number of moles of K2CO3
=mass of K2CO3/molar mass of K2CO3
=2.0/(39.1 x 2+12.0+16.0 x 3)
=0.0145 mol
K2CO3 is the limiting reactant here.
Mole ratio of K2SO4 to K2CO3 = 1:1
Hence,
Number of moles of K2SO4
=0.0145 mol
Mass of K2SO4
=0.0145 x molar mass of K2SO4
=0.0145 x (39.1 x 2+32.1+16.0 x 4)
=2.52 g