"C_3H_8O+4.5O_2\\to3CO_2+4H_2O"
"60\\ g" Propanol produce "44\\times 3=132\\ g" of "CO_2"
"100 \\ Kg" Propanol will produce "\\frac{132}{60}\\times100\\ Kg =220\\ Kg" of "CO_2"
"60\\ g" Propanol consume "4.5\\times 32=144 \\ g" of "O_2"
"100 \\ Kg" Propanol consume "\\frac{144}{60}\\times 100 \\ Kg=240\\ Kg" of "O_2" .
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