## Fundamentals M7

Camellia Liu 1J
Posts: 51
Joined: Sat Aug 24, 2019 12:15 am

### Fundamentals M7

I found that Mg was the limiting reactant and that B2O3 was in excess by using the ratio 1:3 and finding that since we have approx 1795 moles of B2O3, we would need approx 5385mol of Mg. Since we only have 5143 mols of Mg, I concluded that it was the limiting reactant. Is there a faster/ more clear-cut way of finding the limiting and excess reactants? Are my answers/conclusions correct? My final answer for part b was approx 37.063 kg B produced and I was wondering if anyone else also got this answer.

LReedy_3I
Posts: 51
Joined: Fri Aug 30, 2019 12:17 am

### Re: Fundamentals M7

I found the limiting reactant in the same way, I don't know of a faster way than using the molar masses to convert to mols. I also had 37.06 kg of B as the product of the reaction, using the 2:3 molar ratio and the molar mass of Boron.

Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

### Re: Fundamentals M7

Another way of approaching limiting reactant problems is by taking the amount of each reactant in grams and converting it to moles. Afterwards, compare the ratio of moles present:moles required for each reactant. Whichever one has the lower ratio is the limiting reactant. So in this case, 125 kg of B2O3 = 1795 mol B2O3 and 125 kg of Mg = 5142 mol Mg. Compare the ratios of calculated:required and you'll see that:
1795 mol B2O3 / 1 mol B2O3} > 5142 mol Mg / 3 mol Mg
Therefore, Mg is the limiting reactant.

Kate Osborne 1H
Posts: 102
Joined: Fri Aug 30, 2019 12:16 am

### Re: Fundamentals M7

The way I was taught to find the limiting reactant is to convert from grams to moles and then to divide the amount of moles you find by the coefficient in the reaction and then compare those numbers so you do not have to go about comparing ratios and the math gives you a more clear answer as to which is the limiting reactant.