Question #45702

1 Answer
Feb 10, 2016

Answer:

Solution

Explanation:

#H_2 + O_2 \rightarrow H_2O#

You have 2 atoms of O on the left while you have one on the right side of the equation. Correct that as below

#H_2 + O_2 \rightarrow 2H_2O#

Now oxygen molecule is balanced. Take hydrogen. 2 atoms on the left side. We have 4 on the other. Balance this.

#2H_2 + O_2 \rightarrow 2H_2O#

Now all the elements are balanced in the equation.

As per this equation 2 moles of #H_2 # and 1 mole of #O_2# give 2 moles of water.

1 gmole of #O_2# = 32 g of #O_2#
10 g of #O_2# = #10/32# gmole of #O_2#
# = 0.3125# gmole of #O_2#.

If we have #H_2# available at a quantity #\ge 2 \times 0.3125# gmole (i.e) #0.625# gmole then the product formed will be
#H_2O# formed# = 2 \times 0.3125# gmole
Molecular weight of water #=18# g/gmole
#H_2O# formed# = 2 \times 0.3125 * 18 # g/gmole *gmole
#H_2O# formed# = 11.25# g.

Otherwise the product formed is a function of #H_2# availability.