Question

Please Help. Will Give Medal!!!
Suppose you had 0.70 g of Ba(NO3)2 and 1.24 g of NH2SO3H, which would be the limiting reactant in reactions (a), (b) and (c)? Is the limiting reactant the same for all three?

a. _____ Ba(NO3)2(aq) + _2_ NH2SO3H(aq) -> _____ Ba(NH2SO3)2(s) + _2_ HNO3(aq)

b. _____ Ba(NO3)2(aq) + _____ NH2SO3H(aq) + ____ H2O(l) -> ____ BaSO4(s) + ____ NH4NO3(aq) + ___ HNO3(aq)

c. ____ Ba(NO3)2(aq) + _2_ NH2SO3H(aq) + _2_ H2O(l) -> ____ Ba(NH2)2(s) + _2_ H2SO4(aq) + _2_ HNO3(aq)

Answer 1

the answer is in here: http://plato.mercyhurst.edu/chemistry/rbrown/WWW/gc2lab/Winter2011-12%20GenChem%20II%20Lab%20Manual.pdf

Answer 2

Only the question is there...

Answer 3

you have to calculate how many moles of each of the reactants you have in the given grams, then look at the relationship of the reactants in the 3 reactions, the a) is 1:2, b) 1:1 and c)is 1:2.

Answer 4

i dont understand how to convert to moles

Answer 5

number of moles = mass/ molecular mass
you have to calculate the molecular mass of your compounds
do you know how to calculate molecular mass?

Answer 6

no

Answer 7

do you have a periodic table in your hands?

Answer 8

yes

Answer 9

find there the atomic mass of all the elements that compose your first compound, Ba, N, O
Ba(NO3)2

Answer 10

okay. then what do i do with those numbers?

Answer 11

the atomic mass is the 4-5 digit number in the box of the element
for Ba = 137.327, N= 14.0067 , O= 15.9994

Answer 12

you are good in math!
you have to add this number as the proportions in the formula

Answer 13

1 Ba + 2 x (1N +3 x O) =

Answer 14

im confused

Answer 15

137.327+ 2 (14.0067 +( 3 x 15.9994))= 261.337

Answer 16

what is the 261.337? i'm sorry, i just don't understand this problem.

Answer 17

that is the molecular mass of Ba(NO3)2

Answer 18

it is equal to the mass of one mol of Ba(NO3)2 in grams

Answer 19

okay. so i find the molar mass of Ba(NO3)2 and NH2SO3H, then what do i do?

Answer 20

then calculate how many mol of each compound do you have in the 0.70 g of Ba(NO3)2 and 1.24 g of NH2SO3H

Answer 21

how do i do that?

Answer 22

you have to divide the "mass/molecular mass" 0.70/261.337=0.002678

Answer 23

okay

Answer 24

for the reaction a) you need one mole of Ba(NO3)2 every 2 mol of NH2SO3H 1:2 relationship

Answer 25

okay. and is Ba(NO3)2 the limiting reactant in all 3?

Answer 26

and you have a 1:5 relationship 0.002678/0.01277, that means the you have excess of NH2SO3H and your limiting reactant is Ba(NO3)2

Answer 27

I think so...

Answer 28

okay thank you for helping me!