Answer:
2
Explanation:
Mass of water molecule = mass of hydrated salt - mass of anhydrous salt
Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.
Number of moles = mass / molarmass
Molar mass of water = 18.015g/mol
No. of moles of water = 0.74 / 18.015 = 0.0411 moles.
Mass of BaCl2 present =?
1 mole of BaCl2 = 208.23 g
X mole of BaCl2 = 4.26 g
X = (4.26 * 1) / 208.23
X = 0.020
0.020 moles is present in 4.26g of BaCl2
Mole ratio between water and BaCl2 =
0.0411 / 0.020 = 2
Therefore 2 molecules of water is present the hydrated salt.
Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:

= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
Answer : The mole fraction of nitrogen will be 0.4615.
Explanation : When nitrogen (
)and hydrogen (
)are mixed, the mole ratio becomes 1 : 1.5,
Now we know that (
) is acting as a limiting agent.
So at the time of when 0.4 moles of (
) is been formed it requires 0.4 moles of (
) and 3.4 moles of (
)
So, we find the the remaining (
) will be 0.6 and
(
) will be 0.3 mole present in mixture.
So, the mole fraction of (
) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615
Convert moles to mass.
mass C = 0.2 mol * 12 g / mol = 2.4 g
mass H = 0.4 mol * 1 g / mol = 0.4 g
So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g
Calculating for moles O given mass:
moles O = 3.2 g / (16 g / mol) = 0.2 moles
Answer:
<span>0.2 moles O</span>