What is the theoretical mass of water produced from ten moles of #NaHCO_3#?

1 Answer
Meave60
Aug 11, 2018

The theoretical mass of water that can be produced by the decomposition of ten moles of sodium bicarbonate is #"90 g"#.

Explanation:

Balanced equation

#"2NaHCO"_3("s")##rarr##"Na"_2"CO"_3("s") + "CO"_2("g") + "H"_2"O(g)"#

You first need to determine the number of moles of water produced, then multiply the moles of water produced by its molar mass, #"18.105 g/mol"#.

To calculate the moles of water produced, multiply the moles of #"NaHCO"_3"# by the mole ratio between #"NaHCO"_3"# and #"H"_2"O"# in the balanced equation, with moles #"H"_2"O"# in the numerator.

#10color(red)cancel(color(black)("mol NaCO"_3))xx(1"mol H"_2"O")/(2color(red)cancel(color(black)("mol NaHCO"_3)))="5 mol H"_2"O"#

Calculate the mass of #"H"_2"O"# by multiplying the moles by its molar mass.

#5color(red)cancel(color(black)("mol H"_2"O"))xx(18.015"g H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O")))="90 g H"_2"O"#

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