Question

2. For the following reaction, what volume of 1.8M Na2CO3 is needed to react with 10.0L of 0.30M of

AICI3

Answer 1

Answer:

Molarity = 3.33 L

Step-by-step explanation:

The balanced chemical equation for the reaction is:

Na2CO3 + 2AlCl3 → 2NaCl + Al2(CO3)3

To determine the volume of 1.8 M Na2CO3 needed, we need to use the stoichiometry of the reaction and the formula:

Molarity (M) = moles of solute / liters of solution

First, let's find the moles of AlCl3 in 10.0 L of 0.30 M solution:

moles AlCl3 = Molarity x Volume = 0.30 mol/L x 10.0 L = 3.0 mol

From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of AlCl3. Therefore, we need:

moles Na2CO3 = (2/1) x moles AlCl3 = 2 x 3.0 mol = 6.0 mol

Now we can use the formula for molarity to find the volume of 1.8 M Na2CO3 needed:

Molarity = moles of solute / liters of solution

1.8 mol/L = 6.0 mol / liters of Na2CO3

Solving for liters of Na2CO3:

liters of Na2CO3 = moles of solute / Molarity = 6.0 mol / 1.8 mol/L = 3.33 L

Therefore, we need 3.33 L of 1.8 M Na2CO3 to react with 10.0 L of 0.30 M AlCl3.