Answer : The mass of
required are, 35 kg
Explanation :
First we have to calculate the mass of
.
The first step balanced chemical reaction is:
![2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2](https://tex.z-dn.net/?f=2MnCO_3%2BO_2%5Crightarrow%202MnO_2%2B2CO_2)
Molar mass of
= 115 g/mole
Molar mass of
= 87 g/mole
Let the mass of
be, 'x' grams.
From the balanced reaction, we conclude that
As,
of
react to give
of ![MnO_2](https://tex.z-dn.net/?f=MnO_2)
So,
of
react to give
of ![MnO_2](https://tex.z-dn.net/?f=MnO_2)
And as we are given that the yield produced from the first step is, 65 % that means,
![60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg](https://tex.z-dn.net/?f=60%5C%25%20%5Ctext%7B%20of%20%7D0.757xg%3D%5Cfrac%7B60%7D%7B100%7D%5Ctimes%200.757x%3D0.4542xg)
The mass of
obtained = 0.4542x g
Now we have to calculate the mass of
.
The second step balanced chemical reaction is:
![3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3](https://tex.z-dn.net/?f=3MnO_2%2B4Al%5Crightarrow%203Mn%2B2Al_2O_3)
Molar mass of
= 87 g/mole
Molar mass of
= 55 g/mole
From the balanced reaction, we conclude that
As,
of
react to give
of ![Mn](https://tex.z-dn.net/?f=Mn)
So,
of
react to give
of ![Mn](https://tex.z-dn.net/?f=Mn)
And as we are given that the yield produced from the second step is, 80 % that means,
![80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg](https://tex.z-dn.net/?f=80%5C%25%20%5Ctext%7B%20of%20%7D0.287xg%3D%5Cfrac%7B80%7D%7B100%7D%5Ctimes%200.287x%3D0.2296xg)
The mass of
obtained = 0.2296x g
The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)
So, 0.2296x = 8000
x = 34843.20 g = 34.84 kg = 35 kg
Therefore, the mass of
required are, 35 kg