Chemistry question

Question

In the reaction of 4.3 g of h2 and 6.11g of o2 2.0g of water was isolated. what in the percentage yield of water?

Answer 1

First, you should always use appropriate symbols for chemical reactions.  h2 and o2 are not appropriate.  They are H2 and O2 respectively.

 

2H2(g) + O2(g) ==>2 H2O(g)

moles H2 used = 4.3 g H2 x 1 mole/2.0 g = 2.15 moles H2

moles O2 used - 6.11 g O2 x 1 mole/32 g = 0.191 moles O2 

Limiting reactant is O2 since the mole ration of H2:O2 is 2:1 and 2.15 moles H2 is > 2x0.191 moles O2

Maximum moles of H2O that can be formed = theoretical yield = 0.191 moles O2 x 2 moles H2O/mole O2 = 0.382 moles

Theoretical mass H2O = 0.382 moles H2O x 18 g/mole = 6.87 g H2O

% yield = 2.0 g/6.87 g (x100%) = 29% (to 2 significant figures)