Question

what is the molar mass of 3Na2CO3

Answer 1

This is kind of a funky question because the 3 out in front of the compound means 3 moles, but molar mass is, by definition, per 1 mole.

Answer 2

So, I think you can just find the molar mass of Na2CO3.

Answer 3

You can do that by adding up the atomic masses of each of the elements, multiplied by how many of that element you have in 1 compound.

Answer 4

ah sorry. here is the original question: If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent? (2 points)

Answer 5

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

Answer 6

Ah, I see. I would use conversions to find how much of 1 of the products you can make with each amount of reactant you have.

Answer 7

The reactant that makes less of the same product will run out first, or be the limitingreactant.

Answer 8

yep. i was finding the masses of each

Answer 9

and ratios

Answer 10

Sounds good. :)

Answer 11

but I'm not sure of the mass of 3Na2CO3

Answer 12

the answer is 341 but i can't get that

Answer 13

I don't get that either...

Answer 14

I get the molar mass to be 105.98g/mol

Answer 15

i get 236

Answer 16

so the mole ratios are 2, 3, 1, and 6

Answer 17

and the molar mass of 2Al(NO3)3 is 240.034

Answer 18

To find the molar mass, you just use the subscript numbers.
2 Na, 1 C, and 3 O.

Answer 19

For. Na2CO3.

Answer 20

You can multiply that up to find the mass of 3 moles of the compound, but that is not the molar mass.

Answer 21

ah okay

Answer 22

:)

Answer 23

okay i really need help. so i figured out that

Answer 24

7.00 moles of Na2CO3 4.00 moles of Al(NO3)3 are needed

Answer 25

Al(NO3)3 : k1 = n(Al(NO3)3) / 2 = 4,00 / 2 = 2,00

Na2CO3 : k2 = n(Na2CO3) / 3 = 7,00 / 3 = 2,33
OR
7 mol of Na2CO3 times 2 mole of Al(NO3) divided by 3 mol of NA2COS

Answer 26

All your calculations look great to me! So, what does it mean?