Answer :

1) Balance the chemical equatio.

[tex]Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2[/tex]

2) Which is the limiting reactant.

2.1- How many moles of HCl d we need to use all of the Na2CO3?

The molar ratio between HCl and Na2CO3 is 2 mol HCl: 1 mol Na2CO3.

[tex]mol\text{ }HCl=5\text{ }mol\text{ }Na_2CO_3*\frac{2\text{ }mol\text{ }HCl}{1\text{ }mol\text{ }Na_2CO_3}=10\text{ }mol\text{ }HCl[/tex]

We need 10 mol HCl and we have 8 mol HCl. We do not have enough HCl. This is the limiting reactant.

2.2- How many moles of Na2CO3 do we need to use all of the HCl?

The molar ratio between HCl and Na2CO3 is 2 mol HCl: 1 mol Na2CO3.

[tex]mol\text{ }HCl=8\text{ }mol\text{ }HCl*\frac{1\text{ }mol\text{ }Na_2CO_3}{2\text{ }mol\text{ }HCl}=4\text{ }mol\text{ }Na_2CO_3[/tex]

We need 4 mol Na2CO3 and we have 5 mol Na2CO3. e have ebnough Na2CO3. This is the excess reactant.

3) Moles of NaCl produced.

Limiting reactant: 8 mol HCl.

The molar ratio between HCl and NaCl is 2 mol HCl: 2 mol NaCl.

[tex]mol\text{ }NaCl=8\text{ }mol\text{ }HCl*\frac{2\text{ }mol\text{ }NaCl}{2\text{ }mol\text{ }HCl}=8\text{ }mol\text{ }NaCl.[/tex]

4) Grams of NaCl produced.

The molar mass of NaCl is 58.44 g/mol.

[tex]g\text{ }NaCl=8\text{ }mol\text{ }NaCl*\frac{58.44\text{ }g\text{ }NaCl}{1\text{ }mol\text{ }NaCl}=467.52\text{ }g\text{ }NaCl[/tex]

d. 467.52 g NaCl can be produced.

5) Grams of excess reactant react and get used.

In step 2.2 we got the result below

We need 4 mol Na2CO3 and we have 5 mol Na2CO3. We have enough Na2CO3. This is the excess reactant.

According to this resu4t, mol Na2CO3 reacts and gets used up and 1 mol remains in excess.

5.1- Convert moles to grams (reacted)

4 mol Na2CO3.

The molar mass of Na2CO3 is 105.99 g/mol

[tex]g\text{ }Na_2CO_3=4\text{ }mol\text{ }Na_2CO_3*\frac{105.99\text{ }g\text{ }Na_2CO_3}{1\text{ }mol\text{ }Na_2CO_3}=423.96\text{ }g\text{ }Na_2CO_3[/tex]

e. 423.96 g react and get used up.

5.2-Convert moles to grams (excess)

1 mol Na2CO3.

The molar mass of Na2CO3 is 105.99 g/mol

[tex]g\text{ }Na_2CO_3=1\text{ }mol\text{ }Na_2CO_3\frac{105.99\text{ }g\text{ }Na_2CO_3}{1\text{ }mol\text{ }Na_2CO_3}=105.99\text{ }g\text{ }Na_2CO_3[/tex]

f. 105.99 g of the excess reactant remains in excess.

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