How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion?

#B_2H_6 + O_2 -> HBO_2 + H_2O#

Balanced: #B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#

Molar mass of #B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#

Molar mass of #H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#

Moles of #B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#

From the balanced equation 0.6217 mol of #B_2H_6# produces #2(0.6217) = 1.24 " mol" # of #H_2O#

#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O# (rounded to two significant figures)

Balanced equation

#"B"_2"H"_6("g") + "3O"_2("g")"##rarr##"B"_2"O"_3("s") + "3H"_2"O(g)"#

We first need to determine mol #"B"_2"H"_6"# by dividing its given mass by its molar mass #("25.652 g/mol")#. Divide by multiplying by the reciprocal of its molar mass (mol/g).

#17.2color(red)cancel(color(black)("g B"_2"H"_6))xx(1"mol B"_2"H"_6)/(25.652color(red)cancel(color(black)("g B"_2"H"_6)))="0.671 mol B"_2"H"_6"#

To determine mol #"H"_2"O"# produced, multiply mol #"B"_2"H"_6"# by the mol ratio between #"H"_2"O"# and #"B"_2"H"_6"# from the balanced equation, with #"H"_2"O"# in the numerator.

#0.671color(red)cancel(color(black)("mol B"_2"H"_6))xx(3"mol H"_2"O")/(1color(red)cancel(color(black)("mol B"_2"H"_6)))="2.01 mol H"_2"O"#

The complete combustion of #"17.2 g B"_2"H"_6"# produces #"2.01 mol H"_2"O"#.