A) identify the limiting reagent and predict how many grams of sodium sulfate ar

Question

A) identify the limiting reagent and predict how many grams of sodium sulfate are produced when 2.0g of sulfur reacts with 3.0g of oxygen and 4.0g of sodium hydroxide according to the following chemical equation

2 S(s)+3O2(g)+4NaOH(aq)-----> 2 Na2SO4(aq)+ 2H2O(l)

B) The concentration of chlroide ion (CI-) in an aqueus salt solution can be determined by titrating with a solution of AgNO3-. In this Titration, the following reaction occurs:

AgNO3(aq)+NaCl(aq)--->AqCl(s)+NaNO3(aq)

1) please write out the ionic equations for this reaction

2) What is the concentraion of Cl- (aq) in a 100,0 mL sample of a salt solution if 24,1 mL of 0.100 M AgNO3(aq) are needed to react with all the Cl-(aq) in sample?

please post the answer and explanation so it can be copy and paste into a word document.

Explanation / Answer

A) bal. rxn.: S + O2 + 2 NaOH ------> Na2SO4 + H20
moles of reactants-
S: 2.0/32 g/mole S= 0.063 moles S
O2: 3.0 g /32 g/mole O2=0.094moles O2
NaOH: 4.0g/40 g/mole NaOH= 0.10 moles NaOH

Theoretical yield of Na2SO4 based upon quantities of reactants
S: 0.063 moles S x 142.04 g/mole Na2SO4= 8.95 g Na2SO4
O2: 0.094 moles O2 x 142.04 g/mole Na2SO4= 13.35 g Na2SO4
NaOH: 0.10 moles x 1/2* x 142.04 g/mole Na2SO4= 7.10 g Na2SO4
* from bal. rxn., 2 moles of NaOH reacts to form 1 mole Na2SO4
therefore NaOH is the limiting reactant

B)

AgNO3(aq) + NaCl(aq)====>AgCl(s) + NaNO3(aq)

It's balanced. Now you should realize that this is going to be a double replacement reaction because you have two compounds on the reactant side, AgNO3 and NaCl. A single replacement reaction is when you have an element like Na and a compound like MgCl2.

So since it's a double replacement we switch the place of the positive ions. Ag and Na are metals so they are the positive ions. So we switch the place of Ag and put it with Cl to get AgCl and we switch the place of Na and put it with NO3 to get NaNO3.

After we get:

AgNO3(aq) + NaCl(aq)====>AgCl(s) + NaNO3(aq)

We know our reactants are always aqueous(aq) in a double replacement reaction and you have to memorize the solubility rules or if you're allowed you can use a solubility chart/table to figure out what will be Solid(s) or aqueous(aq) and there are some things that you need to know are gas(g) and liquid(l) such as CO2 is a gas and water is usually liquid in these type of equations. Scroll down below to see some websites that explain the solubility rules.

Anyways once we get:

AgNO3(aq) + NaCl(aq)====>AgCl(s) + NaNO3(aq)

We now know the state of everything. The total ionic is with all the ions. Ions are things that are (aq).

We break apart everything into the ions that it is composed of. For instance AgNO3 is made up of Ag+ and NO3^- so that's what we're going to break up AgNO3 into. NaCl is made up of Na+ and Cl- so that's what we're going to break it up into. NaNO3 is made up of Na+ and NO3^- so that's what we break that up into. Now AgCl is a solid so we don't break that up in the total ionic or net ionic equation. That's one of your mistakes.

Total ionic:
Ag+(aq) + NO3^-(aq) + Na+(aq) + Cl-(aq)====>AgCl(s) + Na+(aq) + NO3^-(aq)

For the net ionic we get rid of everything that is on both sides of the equation in the same state of matter. NO3^-(aq) and Na+(aq) are on both sides of the equation so we get rid of those to get:

Ag+(aq) + Cl-(aq)====>AgCl(s)

That is the net ionic equation. The things that we just got rid of are called spectator ions because they don't actually do anything in the reaction. We just write them in there to make it look nice. So then Na+(aq) and NO3^-(aq) are your spectator ions.

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