## Homework M12

tmehrazar
Posts: 39
Joined: Fri Apr 06, 2018 11:04 am

### Homework M12

For part c on homework problem M12, do we find the excess amount of reactant by subtracting the original amount of Al by the amount of Al we get when we convert Fe0 to grams of Al since FeO was the limiting reactant? If this works, are there other ways to find the excess reactant leftover?

NabilaNizam-1K
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

### Re: Homework M12

Based on the calculations, we can find the masses of the products, Fe and $Al_2O_3$ which are 4.907g and 4.4814 g respectively. Therefore, the mass of excess reactant leftover can be calculated by subtracting the total mass of products from the initial total mass of reactants.

Mass of excess reactants = Initial total mass of reactants (FeO + Al) - Total mass of products (Fe + $Al_2O_3$)

Marisol Sanchez - 1E
Posts: 38
Joined: Fri Apr 06, 2018 11:05 am

### Re: Homework M12

I have not done this problem but having done others of the same nature I think I should be able to clarify. Once you find the limiting reactant, you multiply the amount you have of that reactant (in moles) by the molar ratio you would need of the other reactant in order to find how much you would actually use given the limiting reactant amount.

To do this you multiply the (amount in moles of the limiting reactant you have) by (amount in moles of other reactant/amount in moles of limiting reactant)*

*This second part is relating the molar ratios and is purely based on the chemical equation that is given by the question from the beginning and not of what amounts you actually have of each reactant.

This should give you a value in moles which you can then convert into grams using the molar mass. Once you have accomplished that, you can simply subtract the total amount you have of the reactant by the amount you would actually use and that is the solution to part C.

105012653 1F
Posts: 30
Joined: Fri Apr 06, 2018 11:02 am

^^ agreed