How many grams of barium carbonate can be produced from mixing 100.0 mL of 0.20M barium nitrate and 50.0 mL of 0.30M potassium carbonate?

To calculate how many grams of barium carbonate can be produced, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Let's start by calculating the number of moles of each reactant:

Moles of barium nitrate = (0.20 mol/L) * (0.100 L) = 0.020 mol
Moles of potassium carbonate = (0.30 mol/L) * (0.050 L) = 0.015 mol

Next, we need to determine the molar ratio between barium nitrate and barium carbonate. The balanced equation for the reaction between barium nitrate (Ba(NO3)2) and potassium carbonate (K2CO3) is:

Ba(NO3)2 + K2CO3 → BaCO3 + 2KNO3

From the balanced equation, we can see that the molar ratio between barium nitrate and barium carbonate is 1:1. This means that for every 1 mole of barium nitrate, we can produce 1 mole of barium carbonate.

Since the molar ratio is 1:1 and we have 0.020 moles of barium nitrate, we will be able to produce 0.020 moles of barium carbonate.

Finally, to find the mass of barium carbonate, we need to multiply the number of moles by the molar mass of barium carbonate (BaCO3), which is 197.34 g/mol.

Mass of barium carbonate = 0.020 mol * 197.34 g/mol = 3.95 g

Therefore, approximately 3.95 grams of barium carbonate can be produced from mixing 100.0 mL of 0.20M barium nitrate and 50.0 mL of 0.30M potassium carbonate.