Question

In a double replacement reaction lead(II) nitrate reacts with sodium iodide. Given 5.0 grams of lead (II) nitrate and 3.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

Im really struggling with Stoichiometry! I just dont understand! Please please someone help!

Answer 1

Approximately `1.7` grams of sodium nitrate

Explanation:

Well, we first write down the chemical reaction:

`Pb(NO_3)_2(aq)+2NaI(aq)->PbI_2(s)downarrow+2NaNO_3(aq)`

Note that lead(II) iodide will dissolve only in hot water.

In this problem, we first need to find the limiting reactant. Let's choose a product, such as sodium nitrate in this case.

We need to convert `5 \ "g of" \ Pb(NO_3)_2` and `3 \ "g of" \ NaI` into moles.

Lead(II) nitrate has a molar mass of `331.2 \ "g/mol"`, while sodium iodide has a molar mass of `149.89 \ "g/mol"`. So here, we got:

`(5color(red)cancelcolor(black)"g")/(331.2color(red)cancelcolor(black)"g""/mol")~~0.015 \ "mol"` of lead(II) nitrate

`(3color(red)cancelcolor(black)"g")/(149.89color(red)cancelcolor(black)"g""/mol")~~0.02 \ "mol"` of sodium iodide

Now, let's compare the mole ratios. One mole of lead(II) nitrate produces two moles of sodium nitrate, but two moles of sodium iodide produces two moles of sodium nitrate.

Here,

`0.015color(red)cancelcolor(black)("mol of" \ Pb(NO_3)_2)*(2 \ "mol of" \ NaNO_3)/(1color(red)cancelcolor(black)("mol of" \ Pb(NO_3)_2))=0.03 \ "mol of" \ NaNO_3`

`0.02color(red)cancelcolor(black)("mol of" \ NaI)*(2 \ "mol of" \ NaNO_3)/(2color(red)cancelcolor(black)("mol of" \ NaI))=0.02 \ "mol of" \ NaNO_3`

From here, we clearly see that sodium iodide is the limiting reactant, and all the sodium nitrate produced will depend on the amount of sodium iodide.

So, also `0.02 \ "mol"` of sodium iodide will be formed.

Sodium iodide has a molar mass of `84.9947 \ "g/mol"`.

And so, we would produce:

`0.02color(red)cancelcolor(black)"mol"*(84.9947 \ "g")/(color(red)cancelcolor(black)"mol")~~1.7 \ "g"`