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 answer is
<h2>26.94 g/mL</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 970 g
volume = 36 mL
The density of the object is
We have the final answer as
<h3>26.94 g/mL</h3>Hope this helps you
Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
=
V2 =
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield = = 2.4232 g
percentage of potassium chlorate in the original mixture = = 32.6%