Question

8.0g of certain gas occupies 5.6 L at STP.

A) How many moles of gas are present?
B) What is the molar mass of the gas?
C) What is the common atmospheric gas was collected?

Answer 1

A) Using the ideal gas law, we can calculate the number of moles of gas present:

```

PV = nRT

```

where:

* P = pressure (atm) = 1 atm

* V = volume (L) = 5.6 L

* n = number of moles of gas

* R = ideal gas constant = 0.08206 L atm / mol K

* T = temperature (K) = 273.15 K

Solving for n, we get:

```

n = (P * V) / RT

```

```

n = (1 atm * 5.6 L) / (0.08206 L atm / mol K * 273.15 K)

```

```

n = 0.25 mol

```

Therefore, there are 0.25 moles of gas present.

B) The molar mass of the gas can be calculated by dividing the mass of the gas (8.0 g) by the number of moles of gas (0.25 mol):

```

Molar mass = Mass / n

```

```

Molar mass = 8.0 g / 0.25 mol

```

```

Molar mass = 32 g/mol

```

The molar mass of the gas is 32 g/mol.

C) The common atmospheric gas with a molar mass of 32 g/mol is oxygen (O2). Therefore, the gas that was collected is oxygen.

Step-by-step explanation: