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Chemistry 21 Online
OpenStudy (anonymous):

Please help me with my Chemistry questions!

OpenStudy (kc_kennylau):

That means balance C2H2+O2->CO2+H2O

OpenStudy (kc_kennylau):

LHS: C:2 H:2 O:2 RHS: C:1 H:2 O:3

OpenStudy (kc_kennylau):

\[\begin{array}{rcl} C_2H_2+O_2&\rightarrow&CO_2+H_2O\\ C_2H_2+O_2&\rightarrow&2CO_2+H_2O\\ 2C_2H_2+2O_2&\rightarrow&4CO_2+2H_2O&\mbox{(because of the O in H2O)}\\ 2C_2H_2+5O_2&\rightarrow&4CO_2+2H_2O\\ \end{array}\]

OpenStudy (kc_kennylau):

LHS: C:4 H:4 O:10 RHS: C:4 H:4 O:10

OpenStudy (kc_kennylau):

C2H2:O2 = 2:5 0.700:X = 2:5

OpenStudy (kc_kennylau):

But 0.700:2.5 = 3:10 ...

OpenStudy (kc_kennylau):

No, that means X is not 2.5

OpenStudy (anonymous):

(0.700 L C2H2) x (5 mol 02 / 1 mol C2H2) = 3.50 L 02 Is that correct?

OpenStudy (kc_kennylau):

It's 2 mol instead of 1 mol C2H2

OpenStudy (anonymous):

So (0.700 L C2H2) x (5 mol 02 / 2 mol C2H2) = 3.50 L 02

OpenStudy (kc_kennylau):

so it's not 3.50L

OpenStudy (anonymous):

1.75?

OpenStudy (kc_kennylau):

Yep :)

OpenStudy (anonymous):

(0.700 L C2H2) x (3 mol CO2 / 2 mol C2H2) = 1.05 L CO2 Is that right for the third one?

OpenStudy (kc_kennylau):

It's 4 mol CO2

OpenStudy (anonymous):

(0.700 L C2H2) x (4 mol CO2 / 2 mol C2H2) = 1.40 L CO2

OpenStudy (kc_kennylau):

Yep

OpenStudy (anonymous):

(0.700 L C2H2) x (4 mol H20 / 2 mol C2H2) = 1.40 L H20 vapor

OpenStudy (anonymous):

Is that right for the third one?

OpenStudy (kc_kennylau):

C'est 2 mol H2O

OpenStudy (anonymous):

(0.700 L C2H2) x (2 mol H20 / 2 mol C2H2) = 0.70 L H20 vapor

OpenStudy (kc_kennylau):

Oui c'est bon

OpenStudy (anonymous):

Thank you so much!

OpenStudy (kc_kennylau):

Pas de problème :) Tu m'as fait appris quelque chose aussi :)

OpenStudy (anonymous):

Vous êtes le meilleur!

OpenStudy (anonymous):

I'm unsure about the very last one, hence I didn't include my answer.

OpenStudy (kc_kennylau):

Je ne suis pas sûr, mais cette page web dit non http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=12.8%20L%20of%20helium%20to%20mol

OpenStudy (anonymous):

I am going to post it as a new one. hold on.

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