Hey there!:
Molar mass:
H2 = 2.01 g/mol ; H2O = 18.01
Given the reaction:
2 H2 + O2 = 2 H2O
2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O
mass H2 --------------------- 1.80 g H2O
mass H2 = 1.80 * 2 * 2.01 / 2* 18.01
mass H2 = 7.236 / 36.02
mass H2 = 0.2008 g
Hope that helps!
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
It has 8 bonding electrons
The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
