Question

exactly 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?

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.

Answer 2

Answer: Calcium (A)

Explanation: Instead of using the ideal gas law, you can simply use $n=\frac{v}{24}$ because it tells you that the volume of one mole of gas is 24$dm^{3$. Convert 300$cm^{3}$ into $dm^{3}$ to get 0.3$dm^{3}$. Find the number of moles by calculating $\frac{0.3}{24}$, which gives $\frac{1}{80}$. Then, you know that the equation formed would be:

(M represents a metallic element)

$2M + O_{2} -> 2MO$

From there, to find the number of moles of the element, multiply $\frac{1}{80}$ by 2 (because the mole ratio of oxygen to M is 1:2). You will get $\frac{1}{40}$. Then, use $n= \frac{m}{Mr}$ and rearrange to make Mr the subject:

$Mr = \frac{m}{n}$

Since you know the mass is 1g and the number of moles is $\frac{1}{40}$, substitute both these values in. You will get:

$Mr = \frac{1}{\frac{1}{40} }$

Plug that into the calculator and your Mr is 40. This is very close to 40.1, the data booklet value for the Mr of copper :)