Answer :
By Dividing the moles of the reactants by the moles of the experiments.
Let's the chemical equation: [tex]2Al + 3I_2 --\ \textgreater \ 2AlI_2[/tex]
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.
[tex]1.20g Al * \frac{1mol Al}{26.98g Al} = 0.04448 mol Al[/tex]
[tex]2.40 * \frac{1 mol I_2}{253.8g I_2} = 0.009456 mol I_2[/tex]
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:
[tex] \frac{0.4448}{0.009456} [/tex] = [tex] \frac{1}{4.70} [/tex]
So, the experimental molar ration of the experiment I2 is [tex] \frac{1}{4.70} [/tex]
Hope this Helps :)
Let's the chemical equation: [tex]2Al + 3I_2 --\ \textgreater \ 2AlI_2[/tex]
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.
[tex]1.20g Al * \frac{1mol Al}{26.98g Al} = 0.04448 mol Al[/tex]
[tex]2.40 * \frac{1 mol I_2}{253.8g I_2} = 0.009456 mol I_2[/tex]
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:
[tex] \frac{0.4448}{0.009456} [/tex] = [tex] \frac{1}{4.70} [/tex]
So, the experimental molar ration of the experiment I2 is [tex] \frac{1}{4.70} [/tex]
Hope this Helps :)
