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:
The percent yield of a reaction is 48.05%.
Explanation:
Volume of water obtained from the reaction , V= 5.76 mL
Mass of water = m = Experimental yield of water
Density of water = d = 1.00 g/mL
Theoretical yield of water : T
Moles of tungsten(VI) oxide =
According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:
Mass of 0.6660 moles of water:
0.666 mol × 18 g/mol = 11.988 g
Theoretical yield of water : T = 11.988 g
To calculate the percentage yield of reaction , we use the equation:
The percent yield of a reaction is 48.05%.
Answer:
A. 0.35 M
Explanation:
Hello,
In this case, given the volume and concentration of lithium hydroxide and the volume of chloric acid, we can compute the concentration of the neutralized acid by using the following equation:
Therefore, answer is A. 0.35 M.
Regards.