Question

Upon balancing the equation below, how many moles of sodium cyanide are needed to react completely with 3.6 moles of sulfuric acid? H2SO4 + NaCN yields HCN + Na2SO4 1.2 mol NaCN 1.8 mol NaCN 3.6 mol NaCN 7.2 mol NaCN

Answer 1

oy this is way way back ....

Answer 2

6(H2SO4) + Y(NaCN)

Answer 3

well, 3.6 not 6

Answer 4

Wait what?

Answer 5

you have 3.6 moles of H2SO4, and need some amount of NaCN, Y

3.6(H2SO4) + Y(NaCN)

i cant really make out the resultants tho

Answer 6

the setup as is is just a memory ... so it might be slightly wrong

Answer 7

So you think it's 3.6?

Answer 8

no, im just trying to recall some stuff ....
\[ H_2~S~O_4 + Na~C~N ~\to~ H~C~N + Na_2~S~O_4\]
we have 2 Nas on the left that need to be accounted for on the right
\[ H_2~S~O_4 + 2(Na~C~N) ~\to~ H~C~N + Na_2~S~O_4\]

Answer 9

we have 2 Hs on the left so we would need to have 2 Hs on the right as well

\[H_2~S~O_4 + 2(Na~C~N) ~\to~ 2(H~C~N) + Na_2~S~O_4\]
this looks to balance out all the parts

Answer 10

we have a 1:2 ratio of H2SO4:NaCN

Answer 11

so....

Answer 12

sooo, for every 1 part of H2SO4, you need 2 parts of NaCN ...

Answer 13

if you have 3.6 parts of the one, how many parts do you need of the other?

Answer 14

7.2?

Answer 15

correct

Answer 16

are you 100% sure that is the answer?

Answer 17

if you have your doubts, look thru my work and see if you spot any errors in it ....

Answer 18

Alright, thank you so much!

Answer 19

to balance a chemical reaction; you have to have the same number of parts of elements on each side.

Since we have the same number of parts of elements on each side in a 1 to 2 ratio ... im fairly confident :)

Answer 20

okay thank you so much, can you help me with another question?