Question

ANSWER THE 3 FOLLOWING QUESTIONS WITH EVIDENCE (40 POINTS):

1. What volume of 0.50M CaCl2 can be produced by dissolving 167.5g of CaCl2?
2. What volume of gas is contained within 13.5mol of gas?
3. What mass of H2 is present in 31.7L of H2 gas?

Answer 1

To produce 0.50M CaCl2, 167.5g of CaCl2 needs to be dissolved, resulting in a volume of 8.35 L. 13.5mol of gas occupies a volume of 299 L. 31.7 L of H2 gas has a mass of 2.85 g.

Step-by-step explanation:

1. What volume of 0.50M CaCl2 can be produced by dissolving 167.5g of CaCl2?
To find the volume, we need to use the formula:
Volume = mass / molar mass/concentration.
The molar mass of CaCl2 is 40.08 g/mol. Plugging in the values, we get:
Volume = 167.5 g / 40.08 g/mol / 0.50 mol/L = 8.35 L.
Therefore, 8.35 L of 0.50M CaCl2 can be produced.

2. What volume of gas is contained within 13.5mol of gas?
To find the volume, we need to use the ideal gas law formula:
PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
Assuming standard temperature and pressure (STP, 1 atm and 273 K), the gas constant R is 0.0821 L·atm/(mol·K). Plugging in the values, we get:
V = nRT / P = 13.5 mol * 0.0821 L·atm/(mol·K) * 273 K / 1 atm = 299 L.
Therefore, the volume of gas contained within 13.5mol of gas is 299 L.

3. What mass of H2 is present in 31.7L of H2 gas?
To find the mass, we need to use the formula:
Mass = volume * density.
The density of H2 gas at STP is 0.0899 g/L. Plugging in the values, we get:
Mass = 31.7 L * 0.0899 g/L = 2.85 g.

Therefore, the mass of H2 present in 31.7L of H2 gas is 2.85 g.

Answer 2

1. 3.02 liters

2. 302.2 liters

3. 2.83 grams

Step-by-step explanation:

Question 1.

Let's convert the molarity in mass.

Molarity is mol/L, so if we convert the moles to mass, we are talking about a mass, in 1L of solution. (mol . molar mass)

0.5 mol . 110.98 g/m = 55.49 g

Now we can make a rule of three:

If 55.49 grams of chloride are in 1L of solution

then 167.5 g of chloride are in (167.5g . 1L) / 55.49 g = 3.02L

For question 2 and 3, we assume STP, where pressure is 1 atm and T° is 273K. - Standard conditions of temperature and pressure (STP)

Let's apply the Ideal gases law equation

P . V = n . R .T

1 atm . V = 13.5 moles . 0.082 l.atm / mol .K . 273K

V = (13.5 moles . 0.082 l.atm / mol .K . 273K) / 1 atm = 302.2 liters (question 2)

Question 3.

1 atm . 31.7 L = n . 0.082 l.atm / mol.K . 273K

(1 atm . 31.7 L ) / (0.082 l.atm / mol.K . 273K) = 1.41 moles → n

Let's convert moles to mass ( mol . molar mass)

1.41 mol . 2g/mol = 2.83 grams