Catalytic oxidation of isopropanol (propan-2-ol) by oxygen produces $100\;t$ of a commonly used product.
Calculate the volume of isopropanol $ (\rho=0.786\frac{g}{mL})$ that must be treated by counting with a yield of $75\%$!
Product is acetone (propanone) with $M=58\frac{g}{mol}$
$n_{propanone \;obtained}$ $=$ $\frac{10^8}{58}$
If the yield was $100\; \%$, it could be obtained:
$n_{propanone \;possible}$ $=$ $\frac{10^8}{58}\cdot \frac{100}{75}$
As one mole of isopropanol theoretically provides one mole of acetone we have :
$n_{isopropanol}$ $=$ $\frac{10^{10}}{58\cdot75}$
$m_{isopropanol}$ $=$ $\frac{10^{10}\cdot 60}{58\cdot75}$
$V_{isopropanol}$ $=$ $\frac{10^{10}\cdot 60}{58\cdot75\cdot 0.786}$ =
$1.75\cdot 10^8 \; mL$
