The following reaction is performed in a lab: 3K2CO3 (aq) + 2 Al(NO3)3 (aq) + ----> Al2(CO3)3 (s) + 6KNO3 (aq) If 225 mL of 1.50 M aluminum nitrate is added to an excess of potassium carbonate,...

The following reaction is performed in a lab: 3K2CO3 (aq) + 2 Al(NO3)3 (aq) + ----> Al2(CO3)3 (s) + 6KNO3 (aq) If 225 mL of 1.50 M aluminum nitrate is added to an excess of potassium carbonate, how many grams of aluminum carbonate will be produced?

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ncchemist's profile pic

ncchemist | eNotes Employee

Posted on

Let's look at the chemical equation that you wrote.

3 K2CO3 + 2 Al(NO3)3 --> Al2(CO3)3 + 6 KNO3

We can see from the coefficients that 2 moles of aluminum nitrate will produce 1 mole of aluminum carbonate assuming that there is an excess of potassium carbonate.  So first let's find the number of moles of aluminum nitrate involved.

225 mL = 0.225 liters * (1.5 moles/liter) = 0.3375 moles Al(NO3)3

Now convert this to moles of Al2(CO3)3 produced as discussed above.

0.3375 moles Al(NO3)3 * (1 mole Al2(CO3)3 / 2 moles Al(NO3)3) = 0.169 moles of Al2(CO3)3

Now convert this amount into grams.

0.169 moles Al2(CO3)3 * (233.99 g / 1 mole) = 39.54 grams Al2(CO3)3

So 39.54 grams of aluminum carbonate will be produced.

Sources:
ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

Posted on

When given molarity and volume, you can automatically use this equation:

`M = ("mol")/("L")`

Use stoichiometry to convert mL to L.

`225 "mL" = (1 "L")/(1000 "mL") = 0.225 "L"`

Use the first equation:

`1.50 M = "mol"/(0.225 "L")`

`"mol" = 0.3375`

Now that you know the moles of aluminum nitrate present, you can use stoichiometry to find out how many grams of aluminum carbonate will be produced.

`0.3375"mol" = (1 "mol")/(2 "mol") = (233.99 "g")/(1 "mol") = 39.5 "g"`

Make sure your answer is in three sig figs!

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