Question

A sample of magnesium hydroxide, Mg(OH)2, is made by adding an excess of aqueous sodium hydroxide to an aqueous solution containing 1.20 g magnesium sulfate, MgSO4. The mass of magnesium hydroxide formed is 0.26 g. What is the percentage yield of magnesium hydroxide?

Answer 1

Answer:

55.17%

Explanation:

2NaOH+MgSO4---->Mg(OH)2+Na2(SO4)

So 1 mole of mgso4 is supposed to yield 1 mole of mg(oh)2

No of moles of mgso4 used is given by:-

1.2/120=0.01 moles

Since mass of mg(oh)2 is 58, the mass the originally should have been 0.58 gms

But it yields 0.26 gms instead

Percentage is (0.58-0.26)*100/0.58

=55.17%

Hope this answer helmed you

Answer 2

Answer:

44.83

Explanation:

MgSO4 + 2NaOH ==> Mg(OH)2 + Na2SO4

mole of MgSO4 used= 1.2/120 = 0.01

thus mass Mg(OH)2 formed= 0.01 x 58= 0.58g

% = 0.26/0.58 x 100 = 44.83%