You omitted the coefficient of 6 for 6O2 in the balanced equation but that won't change anything.
Your value of 1.76 is ok except for the number of significant figures. You need to correct all of them especially if your prof is picky about s.f. You are given 130.6 g (4 s.f.) so you need to carry at least four through the calculations. I like to carry an extra one (to prevent rounding errors, then round to four at the end). For example, I divided 130.6/74.123 = 1.7619.
Except for s.f. you values of 7.04 for moles CO2 and 8.80 for moles H2O are ok.
But you made error in converting mols to grams. First, you apparently multiplied moles H2O by molar mass CO2 (44.011) and you multiplied moles CO2 by what you thought was molar mass H2O but molar mass of H2O is 18.015 and not 33.008. Correct those and you will be ok.
Your post is VERY difficult to read because you didn't capitalize the beginning letter of a sentence and you insisted on using i when you meant I. Proper punctuation is a fact of life and it's better to join the crowd sooner rather than later. I can tell you, for a fact, that had I written this response in the same way you wrote your answer, my run together sentences would have been useless to you.
From the following combustion: 1-butanol + oxygen+ carbon dioxide + water. How many grams of carbon dioxide and water are produced when 130.6 grams of 1-butanol is completely burned? I wrote out the equation and got C4H10) + 02= CO2 + H2O. i balanced the equation and got C4H10O + O2 = 4CO2 + 5H2O. then i divided 130.6 by 74.12 (the mass of 4c 10h and 1O) to get 1.76moles. then i got
1mol C4H10O/1.76=5 mol H2O/x of water and got 8.8 moles H20. then i got
1mol C4H10O/1.76=4CO2/x of carbon dioxide and got 7.04 moles. i then multipled 8.8 X 33.008 (the mass of CO2) and got 290.47 grams. and 7.04 X 44.01 (the mass of H20) and got 309.83 grams. i then added them together and got 600.3 grams. this seems like a high number. is this correct?
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