Question

If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent?

Answer 1

Hello @math92130 !
Can u write the balanced equation for above reaction ?

Answer 2

yep. 2 Al(NO3)3 + 3Na2CO3 = Al2(CO3)3(s) + 6NaNO3

Answer 3

So, 2 moles of Al(NO3)3 reqcts with 3 moles Na2CO3

Answer 4

so how many grams of Al(NO3)3 reacts with how many grams of Na2CO3 ?

Answer 5

wait shouldn't the mole ratio and molar masses be found

Answer 6

426 grams of Al(NO3)3 reacts with 318 grams Na2CO3
So 852.04 gram will require 636 grams. Given mass of Na2CO3 is 741.93 grams. So Na2CO3 is not the limitting reagent

Answer 7

al(no3)3 is the limiting reagent

Answer 8

yes

Answer 9

Getting it ?

Answer 10

According to the balanced equation,
3 moles of Na2CO3 is required to react with 426 grams of al(no3)3. So for 1 gram na2co3 426/318 gram al(no3)3 is required

Answer 11

Now for 741 g na2co3 426/318 * 741 = 993 gram al(no3)3 is required. But only 852.04 gram is given. So it's a limitting reagent

Answer 12

okay thanks