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dddan
Group Title
in the reaction of Si+2Cl yields SiCl4
in this reaction 0.507 mole of SiCl4 is produced how many moles of molecular chlorine were used in the reaction?
please explain.
 one year ago
 one year ago
dddan Group Title
in the reaction of Si+2Cl yields SiCl4 in this reaction 0.507 mole of SiCl4 is produced how many moles of molecular chlorine were used in the reaction? please explain.
 one year ago
 one year ago

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Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
\[ \text{Si} + 2\text{Cl} \rightarrow \text{SiCl}_4 \]Is this equation balanced? No! Why? Because the problem, as posted, is incorrect. Chlorine gas must be diatomic: Cl_2. After doing that, the equation is balanced. From here, use the coefficient ratios for mole ratios and it should be easy. Note, from the balanced equation, 2 moles of chlorine are required to form 1 mole of SiCl_4. If you still need help, let me know. I'll complete the solution for you. :D
 one year ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
thanks, can you show me?
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Sure. Do you see how 1 mole of SiCl_4 is formed from 2 moles of chlorine gas? Once you see that, you just do .507*2. :)
 one year ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
yup got it. thank you!
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
No problem! :D
 one year ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
how do you balance out \[O _{2}+NO \rightarrow 0_{2}+NO _{2}\]
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
You cannot. The O_2s cancel out and NO > NO_2 is nonsense. Did you mean something else?
 one year ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
yea it O(3)+NO yields O(2)+NO(2)
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.1
Oh. \[ O_3 + NO \rightarrow O_2 + NO_2 \]Yep, the 3 is hard to read as a subscript. There are 3 Os on the LHS and 4 Os on the RHS. Try balancing the Os. Put a 2 on the O3 and 3 on the O2. You get: \[ 2O_3 + NO \rightarrow 3O_2 + NO_2 \]Now, there are 7 Os on the LHS and 8 Os on the RHS. Notice, here, that you can keep trying what you want but keeping the Ns balanced means keeping the Os imbalanced. The equation cannot be balanced. See here: http://www.gregthatcher.com/Chemistry/BalanceChemicalEquations.aspx
 one year ago
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