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

Aug 11, 2018

The theoretical mass of water that can be produced by the decomposition of ten moles of sodium bicarbonate is $\text{90 g}$.

#### Explanation:

Balanced equation

"2NaHCO"_3("s")$\rightarrow$$\text{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, $\text{18.105 g/mol}$.

To calculate the moles of water produced, multiply the moles of $\text{NaHCO"_3}$ by the mole ratio between $\text{NaHCO"_3}$ and $\text{H"_2"O}$ in the balanced equation, with moles $\text{H"_2"O}$ 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 $\text{H"_2"O}$ 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"