Question #fdd52

1 Answer
Feb 23, 2017

Answer:

(a) 4 mol; (b) 5 mol; (c) 2 mol.

Explanation:

We can figure these out by using the molar ratios from the balanced equations.

From octane

#"C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O"#

#"Amount of H"_2"O" = 1 color(red)(cancel(color(black)("mol C"_3"H"_8))) × ("4 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol C"_3"H"_8)))) = "4 mol H"_2"O"#

From butane

#"2C"_4"H"_10 + "13O"_2 → "8CO"_2 + "10H"_2"O"#

#"Amount of H"_2"O" → 1 color(red)(cancel(color(black)("mol C"_4"H"_10))) × ("10 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol C"_4"H"_10)))) = "5 mol H"_2"O"#

From methane

#"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"#

#"Amount of H"_2"O" → 1 color(red)(cancel(color(black)("mol CH"_4))) × ("2 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol CH"_4)))) = "2 mol H"_2"O"#