Respuesta :
Answer : The metal could be Calcium.
Solution : Given,
Mass of metallic element = 1 g
Volume of oxygen = [tex]300Cm^3=0.3L[/tex] [tex](1L=1000Cm^3)[/tex]
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,
[tex]PV=nRT\\n=\frac{PV}{RT}[/tex]
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 = [tex]0.0821Latm/moleK[/tex]
Now put all the given values in this formula, we get the moles of oxygen gas.
[tex]n=\frac{(1atm)\times (0.3L)}{(0.0821Latm/moleK)\times (298K)}=0.01226moles[/tex]
As per question, the reaction is,
[tex]2M+O_2\rightarrow 2MO[/tex]
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 [tex]2\times 0.01226=0.02452[/tex] moles of metallic element
Now we have to calculate the molar mass of metallic element.
[tex]\text{ Molar mass of metallic element}=\frac{\text{ Mass of metallic element}}{\text{ Moles of metallic element}}=\frac{1g}{0.02452moles}=40.783g/mole[/tex]
Therefore, this molar mass is more closer to the molar mass of the calcium. So, this metal could be Calcium and form calcium oxide.
