Chemistry (Please explain)

Question

1.) How much of a 1.80M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 190.0mL of a 0.300M barium nitrate solution?

 

2.) 2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)

A 527-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 19.1g CO2. What was the concentration of the HCl solution?

Answer 1

We first have to know that the formula for sodium sulfate is Na₂SO₄ since sulfate has a charge of -2, and sodium a charge of +1. Barium nitrate has a formula of Ba(NO₃)₂ for similar reasons. The precipitate that we are trying to form is BaSO₄, so for every 1 mole of barium, there must be 1 mole of sulfate. In 190 mL of a 0.3 M (mol/L) solution of barium nitrate, there are:

 

0.19 L × 0.3 mol/L = 0.057 mol

 

So there are 0.057 moles of barium nitrate, and since there is one mole of barium for every one mole of barium nitrate (i.e., the chemical formula shows only one atom of it), then there must be 0.057 moles of barium ions. Since there are that many ions of barium, you must add the same number of moles of sulfate, and to do that, we do:

 

0.057 mol ÷ 1.8 mol/L = 0.03167 L

 

So you must add 31.67 mL of sodium sulfate in order to precipitate all of the barium ions.

 

For number 2, as seen in the given formula, we know that 2 moles an acid reacts with 1 mole of carbonate to form 1 mole of carbon dioxide (along with 1 mole of water and 2 moles of a salt). We know that 19.1 g of carbon dioxide forms, and using the molar mass of carbon dioxide (44.01 g/mol) we can find the number of moles of carbon dioxide that is:

 

19.1 g ÷ 44.01 g/mol = 0.434 mol

 

So now we know that 0.434 moles of carbon dioxide formed. In the formula, we know that for every 1 mole of carbon dioxide, there were 2 moles of HCl that reacted with it, so there was twice the amount of acid as there was CO₂, or 0.868 moles of HCl. To find the molarity of the solution, we divide the moles by the liters:

 

0.868 mol ÷ 0.527 L = 1.647 M