volandam1019
Answered

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how many grams of O2 are required to burn 46.0 grams of C5H12

Sagot :

(MM)C5H12=5x12+12x1=72g/mol
(MM)O2=2x16=32g/mol
Acoording to equation:
C5H12 + 8O2 => 5CO2 + 6H2O

The proportion between C5H12 and O2 is 1:8 in mol

So: 72/16=46/x finale x=10,2g of O2
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