Answer:
1,31÷2 =10,11
Explanation:
c10h22+31÷2o2=10co2+11h2o
Answer:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
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
Metals are ductile,malleable,good conductors and shiny
