Balance the equation and answer in terms of

moles.
Chlorine gas can be produced commercially
by passing an electric current through a con-centrated solution of sodium chloride (brine).
The unbalanced equation is
NaCl(aq) + H2O(`) !
NaOH(aq) + Cl2(g) + H2(g) .
If the brine contains 389 g of NaCl, how
much Cl2 can be produced?
Answer in units of mol

Here is a worked example of a stoichiometry problem. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To balance the equation, we need to make sure that there is an equal number of each type of atom on both sides of the equation.

Step 1: Start by balancing the atoms other than hydrogen and oxygen. In this case, we have sodium (Na) and chlorine (Cl) on the left side and sodium (Na), chlorine (Cl), and oxygen (O) on the right side.

NaCl(aq) + H2O(`) → NaOH(aq) + Cl2(g) + H2(g)

The number of sodium atoms (Na) is already balanced on both sides, so let's move on to chlorine (Cl). There is one chloride (Cl) ion on the left side and two chloride (Cl) atoms on the right side. To balance the chlorine atoms, we need a coefficient of 2 in front of NaCl(aq):

2 NaCl(aq) + H2O(`) → NaOH(aq) + Cl2(g) + H2(g)

Step 2: Now let's balance the hydrogen (H) atoms. There are two hydrogen (H) atoms on the left side and two hydrogen (H) atoms on the right side, so they are already balanced.

2 NaCl(aq) + H2O(`) → NaOH(aq) + Cl2(g) + H2(g)

Step 3: Finally, let's balance the oxygen (O) atoms. There is one oxygen (O) atom on the left side and one oxygen (O) atom on the right side, so they are already balanced.

2 NaCl(aq) + H2O(`) → NaOH(aq) + Cl2(g) + H2(g)

Now that the equation is balanced, we can determine the amount of Cl2 produced when 389 g of NaCl is used.

The molar mass of NaCl is 58.44 g/mol, so we can convert the mass of NaCl to moles:

389 g NaCl × (1 mol NaCl / 58.44 g NaCl) = 6.67 mol NaCl

From the balanced equation, we can see that the stoichiometric ratio between NaCl and Cl2 is 2:1. This means that for every 2 moles of NaCl, we can produce 1 mole of Cl2.

Therefore, the amount of Cl2 produced can be calculated as:

6.67 mol NaCl × (1 mol Cl2 / 2 mol NaCl) = 3.34 mol Cl2

So, when 389 g of NaCl is used, approximately 3.34 moles of Cl2 can be produced.