Answer:
Ooh sweetie goodluck
Explanation:
May the gods ever be in your favor.
Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of butane
b) moles of oxygen
According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
Step one calculate the moles of each element
that is moles= % composition/molar mass
molar mass of Ca = 40g/mol, S= 32 g/mol , O= 16 g/mol
moles of Ca = 29.4 /40g/mol=0.735 moles, S= 23.5/32 =0.734 moles, O= 47.1/16= 2.94 moles
calculate the mole ratio by dividing each mole with smallest mole that is 0.734
Ca= 0.735/0.734= 1, S= 0.734/0.734 =1, O = 2.94/ 0.734= 4
therefore the emipical formula = CaSO4
Answer:
Oxygen is limiting reactant
Explanation:
2 H2 + O2 ======> 2 H2 O
from this equation (and periodic table) you can see that
4 gm of H combine with 32 gm O2
H / O = 4/32 = 1/8
32 /16 = 2/1 shows O is limiter
for 32 gm H you will need 256 gm O and you only have 16 gm
