Answer:
45.02 L.
Explanation:
- Firstly, we need to calculate the no. of moles of water vapor.
- n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
- We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.
<em><u>Using cross multiplication:</u></em>
1.0 mole of CO occupies → 22.4 L.
2.01 mole of CO occupies → ??? L.
∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.
Answer:

Explanation:
In this case, we can start with the reaction:

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:


Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):


The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol
= 48 g
(therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

I hope it helps!
Http://www.lcmrschooldistrict.com/demers/cbphysicalscience/Chp%208-3%20Solubility%20and%20Concentrat...
This website helped me understand the concept.