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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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.

Chemistry
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Any ideas? You'll have to start from 42 grams of \(CO_2\) and relate this to grams of benzene (\(C_6H_6\)).
how do i find grams of benzene @Woodward
There is a "fool-proof" 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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