By Dividing the moles of the reactants by the moles of the experiments.
Let's the chemical equation: \(2Al + 3I_2 --\ \textgreater \ 2AlI_2\)
We know that 1.20 g of Al is reacting with 2.40g of I2
and check the periodic table, and u'll see that the mass of 1 atom of Al is 26.98 and the mass of 1 atom of I2 is 253.8 g
First, you convert every single mass you got into moles.
\(1.20g Al * \frac{1mol Al}{26.98g Al} = 0.04448 mol Al\)
\(2.40 * \frac{1 mol I_2}{253.8g I_2} = 0.009456 mol I_2\)
Now, it's time to find the molar ratio
The mole of the reactant is 0.04448 mol, and the experiments are 0.009456 mol.
You divide reactant over experiments, and you get:
\(\frac{0.4448}{0.009456}\) = \(\frac{1}{4.70}\)
So, the experimental molar ration of the experiment I2 is \(\frac{1}{4.70}\)
Hope this Helps :)
