Question

A student tries to mix copper (II) chloride with sodium nitrate. When that student combines 3 grams of copper (II) chloride with 4 grams of sodium nitrate, what mass of sodium chloride is dissolved in solution?

Im really struggling this unit in chemistry! Please help!

Answer 1

Approximately `2.6` grams.

Explanation:

I'll write down the steps, and this is how I solve these types of problems.

• Write down the balanced chemical reaction

• Convert all the amount of reactants given into moles

• Choose one product, and then use mole ratios from both reactants to see which one produces more of the chosen product. The reactant that produces more product is in excess, and that leaves you with the limiting reactant

• Use the amount of product formed by the limiting reactant, that's the total amount of product that will be formed!

• Convert to mass by multiplying by the product's molar mass if necessary

EXTRA: In some cases, you might be asked to find the amount of excess reactant that is left over. In that case, use mole ratios BETWEEN the two reactants, and the subtract the amount of excess reactant needed to react fully with the limiting reactant from the given amount of excess reactant. Then, multiply by the molar mass to find out the mass that hasn't reacted.
Let's do this then!

Step 1. Balanced chemical equation for this reaction is:

`CuCl_2(aq)+2NaNO_3(aq)->Cu(NO_3)_2(aq)+2NaCl(aq)`

Step 2. We are given `3 \ "g"` of copper(II) chloride and `4 \ "g"` of sodium nitrate. In moles that is,

`(3color(red)cancelcolor(black)("g of" \ CuCl_2))/(134.45color(red)cancelcolor(black)("g of" \ CuCl_2)"/mol")=0.0223 \ "mol"` of copper(II) chloride

`(4color(red)cancelcolor(black)("g of" \ NaNO_3))/(84.9947color(red)cancelcolor(black)("g of" \ NaNO_3)"/mol")=0.047 \ "mol"` of sodium nitrate

Step 3. I'll pick sodium chloride here. The mole ratio between copper(II) chloride and sodium chloride is `1:2`, and the mole ratio between sodium nitrate and sodium chloride is `2:2=1`.

From copper(II) chloride, we produce:

`0.0223color(red)cancelcolor(black)("mol of" \ CuCl_2)*(2 \ "mol of" \ NaCl)/(1color(red)cancelcolor(black)("mol of" \ CuCl_2))=0.0446 \ "mol of" \ NaCl`

From sodium nitrate, we produce:

`0.047color(red)cancelcolor(black)("mol of" \ NaNO_3)*(2 \ "mol of" \ NaCl)/(2color(red)cancelcolor(black)("mol of" \ NaNO_3))=0.047 \ "mol of" \ NaCl`

And so, we see that: `0.0446<0.047`. Therefore, copper(II) chloride is the limiting reactant in this case, and we'll only produce `0.0446 \ "mol"` of sodium chloride.

Step 4. Final step is to find the mass of sodium chloride produced. Multiply by its molar mass:

`0.0446color(red)cancelcolor(black)"mol"*(58.4 \ "g")/(color(red)cancelcolor(black)"mol")=color(red)underlinecolor(black)(2.6 \ "g")`