Solution volume is 2.03 litres.
Explanation:
First number of moles of KNO3 present in 555.0 gm will be calculated.
n= mass of the substance/ mass of one mole of the substance.
n= 555/ 101.1032 ( molecular weight of KNO3 is 101.103)
= 5.4 moles
Now volume can be calculated by the formula:
M= n/V (M is the molarity, n is number of moles, V is the volume.)
= n/M
= 5.4/2.7
V = 2.03 liters.
Molarity is defined as the number of moles of a substance dissolved in 1 litre of solution.
Answer: 51.45 grams of excess reagent is left after the completion of reaction.
Explanation: For the calculation of moles, we use the formula:
....(1)
Given mass = 92 grams
Molar mass = 28g/mol
Putting values in equation 1, we get:
- For
Given mass = 112 grams
Molar mass = 116g/mol
Putting values in equation 1, we get:
The reaction follows:
By Stoichiometry,
2 moles of reacts with 3 moles of silicon
So, 0.965 moles of reacts with = = 1.4475 moles of Silicon.
As, the moles of silicon is more than the required amount and is present in excess.
So, the excess reagent for the reaction is Silicon.
Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles
To calculate the amount of Silicon left in excess is calculated by using equation 1:
Amount of Silicon in excess will be 51.45 grams.
What time tho?
you did not put when
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol