Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:
We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so
Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.
Firstly need to determine the empirical formula of the hydrocarbon. Empirical formula is the simplest whole number ratio of components of the compound. Molecular formula is the actual composition of the components in the compound.
percentage of C - 82.66%
percentage of H - (100-82.66) = 17.34 %
in 100 g of compound ;
mass of C - 82.66 g
mass of H - 17.34 g
C H
mass in 100 g 82.66 g 17.34 g
molar mass 12 g/mol 1 g/mol
number of moles 6.88 mol 17.34 mol
(mass/molar mass)
divide the number of moles by least number of moles (6.88 mol)
6.88 mol/6.88 17.34/6.88
1 2.52
multiply these by 2 to get a whole number
C - 1x 2 = 2
H - 2.52 x 2 = 5.04
round off to nearest whole number
C - 2
H - 5
ratio of C to H is 2:5
empirical formula - C₂H₅
empirical formula mass = 12 g/mol x 2 + 5 * 1 g/mol = 29 g
next have to find how many empirical units are there in the molecular unit
molecular unit mass = 58.12 g
empirical unit = 29 g
then number of empirical units = 58.12 / 29 = 2
rounded off , number of empirical units = 2
(C₂H₅) * 2 units
molecular formula = C₄H₁₀
Explanation:
Its the last answer #""_53∧131"I" #
Explanation:
To a reaction to be occured
The formula is
IE is internal energy of reactant and AE is activation energy and TE is threshold energy .
So option D is correct
The number of protons is the Atomic number