What is the theoretical mass of water produced from ten moles of NaHCO_3NaHCO3?

1 Answer
Aug 11, 2018

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

Explanation:

Balanced equation

"2NaHCO"_3("s")2NaHCO3(s)rarr"Na"_2"CO"_3("s") + "CO"_2("g") + "H"_2"O(g)"Na2CO3(s)+CO2(g)+H2O(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"18.105 g/mol.

To calculate the moles of water produced, multiply the moles of "NaHCO"_3"NaHCO3 by the mole ratio between "NaHCO"_3"NaHCO3 and "H"_2"O"H2O in the balanced equation, with moles "H"_2"O"H2O 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"