Hydrogen react with with carbon monoxide(Carbon (ii) oxide) to form methyl alcohol
CO + 2H2 ----> CH3OH
step one calculate the number of moles of H2
= 11g/2.0 g/mol (molar mass of H2 moles = 5.5moles
calculate the moles CO = 37.25grams/28 =1.330moles
The reacting ratio of of H2 to CH3OH 2:1
1.330 x 2/1+2.661 moles of H2 but there is more H2 present than that of co
therefore the grams of CH3OH is threfore =1.330 x32(molar mass of CH3OH) =42.65grams
0.035 mol CF₄ x 4 mols F/ 1 mol CF₄ = 0.14 mols F
0.14 mols F x 6.022x10²³/ 1 mol F= 8.43x10²² atoms
answer: 8.4×10²² atoms of F
maybe try this im not 100% sure though
Answer: 16.32 g of
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of
require = 1 mole of
Thus 0.34 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of
left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of
as excess reagent are left.
Answer:
Usually you have V, for voltage, Ohms for resistance, and Coulombs for charge.
Explanation:
See the image attached :)