Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer:
1
Explanation:
there is only one sulphur (S) atoms in the formula for sodium sulphate
Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2 2x= +12
x= +6, oxidation number of one chromium is +6.
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
