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anonymous
 one year ago
In the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2?
2C6H6 + 15O2 → 12CO2 + 6H2O
The molar mass of benzene is 78.1074 grams and that of CO2 is 44.01 grams.
anonymous
 one year ago
In the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2? 2C6H6 + 15O2 → 12CO2 + 6H2O The molar mass of benzene is 78.1074 grams and that of CO2 is 44.01 grams.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Any ideas? You'll have to start from 42 grams of \(CO_2\) and relate this to grams of benzene (\(C_6H_6\)).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i find grams of benzene @Woodward

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1There is a "foolproof" process to solve \(all\) stoichiometry problems. Once the equation is written and balanced: 1. Convert what you're given to moles  this is trick sometimes because you can be given a lot of information, it's good to know what is relevant. For your problem you're given mass, so to interconvert between mass and moles, use the relationship: \(\Large \sf n=\dfrac{m}{M_m}\) (equation A) where, \(M_m\)=molar mass, \(m\)=mass, and \(n\)= moles. 2. Use the stoichiometric coefficients and the moles (\(n\)) of each species in a ratio. Set up a ratio using the species of interest, like so: e.g. for a general reaction: \(\sf \large \color{red}{a}A + \color{blue}{b}B \rightleftharpoons \color{green}{c}C\) where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients , \(\sf \dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\) From here you can isolate what you need.  For example: if you have 2 moles of B, how many moles of C can you produce? solve algebraically: \(\sf\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)  3. Solve for moles of unknown (in this case benzene) 4. Convert moles to mass (use equation A again)
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