Question

1.00 g of a metallic element reacts completely with 300 cm3 of oxygen at 298 K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0 dm3. What could be the identity of the metal? A calcium B magnesium C potassium D sodium 11

Answer 1

Answer : The metal could be Calcium.

Solution : Given,

Mass of metallic element = 1 g

Volume of oxygen = $300Cm^3=0.3L$        $(1L=1000Cm^3)$

Temperature of gas = 298 K

Pressure of the gas = 1 atm

First we have to calculate the moles of oxygen gas.

Using ideal gas law,

$PV=nRT\\n=\frac{PV}{RT}$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = gas constant = $0.0821Latm/moleK$

Now put all the given values in this formula, we get the moles of oxygen gas.

$n=\frac{(1atm)\times (0.3L)}{(0.0821Latm/moleK)\times (298K)}=0.01226moles$

As per question, the reaction is,

$2M+O_2\rightarrow 2MO$

From the reaction we conclude that the

1 mole of oxygen react with 2 moles of metallic element

0.01226 moles of oxygen react with $2\times 0.01226=0.02452$ moles of metallic element

Now we have to calculate the molar mass of metallic element.

$\text{ Molar mass of metallic element}=\frac{\text{ Mass of metallic element}}{\text{ Moles of metallic element}}=\frac{1g}{0.02452moles}=40.783g/mole$

Therefore, this molar mass is more closer to the molar mass of the calcium. So, this metal could be Calcium and form calcium oxide.