Lol i made a dumb. Of course the final products must stay there.
In (1) and (2) there is exactly 1 mol of Cl
(g) there. Thus, we can immediately perform (1) + (2)
CFCl
3(g) + Cl
(g) + O
3(g) -> CFCl
2(g) + Cl
(g) + ClO
(g) + O
2(g)Cancel out the Cl on both sides:
CFCl
3(g) + O
3(g) -> CFCl
2(g) + ClO
(g) + O
2(g)This new equation, and equation 3, both have exactly 1 mol of ClO present. Thus we can perform (3) + the above
CFCl
3(g) + O
3(g) + ClO
(g) + O
(g) -> CFCl
2(g) + ClO
(g) + O
2(g) + Cl
(g) + O
2(g)Cancel out the ClO:
CFCl3(g) + O3(g) + O(g) -> CFCl2(g) + 2 O2(g) + Cl(g) (Image removed from quote.)
Hey guys im not sure to do q 15 hehe thanks
This is just your ordinary rigorous calculation.
Write out relevant equations:
A) K
2CO
3 + 2 HCl -> CO
2 + H
2O + 2 KCl
B) KHCO
3 + HCl -> CO
2 + H
2O + KCl
C) Na
2CO
3 + 2 HCl -> CO
2 + H
2O + 2 KCl
D) NaHCO
3 + HCl -> CO
2 + H
2O + KCl
Clearly, 1 mol of reactant _ yields 1 mol of CO
2 in all four equations.
Now, determine the moles of each substance present using n=m/MM:
A) n
CO2 = n
K2CO3 = 7.23 * 10
-3 mol
B) n
CO2 = n
KHCO3 = 1.05 * 10
-2 mol
C) n
CO2 = n
Na2CO3 = 9.43 * 10
-3 mol
D) n
CO2 = n
NaHCO3 = 1.19 * 10
-2 mol
So since n = V/V
M, clearly the answer must be D.
And now for the smart wayHowever obvious it was that the moles of the reactant and the moles of CO
2 were the same varies for each person. But without even writing out the full equation you
should've might've been able to see it. This is because CO
32- is being decomposed down to CO
2 by HCl, which has no carbons in it!
Then, because m=1, we have that the
moles is INVERSELY proportional to the molar mass. Thus the one with the
smallest molar mass (NaHCO
3) will yield the largest moles.