2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
H is 4*10^6 M , OH is 2.5*10^-9 M
We are given with
136 g P4
excess oxygen
The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles
Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5
Answer: option B
Explanation: since nuclear fission involves the decay of larger nuclide into smaller nuclei along with Neutron when it is collide with Neutron.
Example Decay of U-235 into Kr and Ba along with 3 neutrons