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.
When the candle is lit and burn to liquid wax, it still has its originally mass, but in liquid form
In order for a certain reaction to be spontaneous, the change in Gibb's free energy must be negative, that is:
dG < 0
We can calculate dG using:
dG = dH - dS
When we plug in the given values, we see that the sign of dG will never be negative; therefore, the reaction will not be spontaneous under any conditions.
Answer:
3.01 L
Explanation:
V
1
: 2.75L
T
1
:
18
∘
C
V
2
: ?
T
2
:
45
∘
C
If you know your gas laws, you have to utilize a certain gas law called Charles' Law:
V
1
T
1
=
V
2
T
2
V
1
is the initial volume,
T
1
is initial temperature,
V
2
is final volume,
T
2
is final temperature.
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
2.75
291
=
V
2
318
By cross multiplying, we have...
291
V
2
= 874.5
Dividing both sides by 291 to isolate
V
2
, we get...
V
2
= 3.005...
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
V
2
= 3.01 L
