Calculate the amount of water formed when #"8 g"# of #"H"_2"# is allowed to react completely with oxygen?

Balanced equation

#"2H"_2 + "O"_2"##rarr##"2H"_2"O"#

#color(red)(1.# Calculate mol #"H"_2"# by dividing its given mass by its molar mass #("2 g/mol")#. Do this by multiplying by the inverse of the molar mass.

#color(blue)(2.# Calculate mol #"H"_2"O"# by multiplying mol #"H"_2"# by the mol ratio between #"H"_2"O"# and #"H"_2"# from the balanced equation, with #"H"_2"O"# in the numerator.

#color(green)(3.# Calculate mass #"H"_2"O"# by multiplying by the molar mass of #"H"_2"O"# #("18 g H"_2"O")#.

#color(red)8color(black)cancel(color(red)("g H"_2))xx(color(red)(1color(black)cancel(color(red)("mol H"_2))))/(color(black)cancel(color(red)(2"g H"_2)))xx(color(blue)2color(black)cancel(color(blue)("mol H"_2"O")))/(color(blue)2color(black)cancel(color(blue)("mol H"_2)))xx(color(green)18color(green)("g H"_2"O"))/(color(green)1color(black)cancel(color(green)("mol H"_2"O")))="70 g H"_2"O"# (rounded to one significant figure)

The equation for the reaction of hydrogen gas and oxygen gas to form water is: #2H_2(g) + O_2(g) -> 2H_2O (l)#

So for every mole of #H_2# you will form a mole of water, assuming you have sufficient oxygen to react with it all.

The atomic weight of hydrogen atoms is 1, but the molar mass of diatomic #H_2# is 2 g/mol. Therefore, 8 g of #H_2# is 4 moles. Therefore you will form 4 moles of water. The molar mass of water is 18 g/mol so you will form 4 x 18 = 72 g of water.