# Given the equation: Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 Li2SO4, how many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? Thanks lots.?

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#### Explanation:

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Meave60 Share
Nov 14, 2015

If this reaction were to occur, $\text{310 g LiNO"_3}$ would be needed to make $\text{250 g Li"_2"SO"_4}$

#### Explanation:

Balanced Equation
$\text{Pb(SO"_4)_2("aq") + "4LiNO"_3("aq")}$$\rightarrow$$\text{Pb(NO"_3)_4("aq")" + 2Li"_2"(SO"_4)("aq")}$

This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.

This reaction will not actually occur. All you will have is a mixture of $\text{Pb"^(4+)}$ ions, $\text{SO"_4""^(2-)}$ ions, $\text{Li"^(+)}$ ions, and $\text{NO"_3""^(-)}$ ions.

However, I will work it out so you can see how a problem of this type is to be done.

Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
You can calculate it or find it in a resource.
$\text{LiNO"_3} :$$\text{68.9459 g/mol}$ http://pubchem.ncbi.nlm.nih.gov/compound/Lithium_nitrate
$\text{Li"_2"SO"_4} :$$\text{109.9446 g/mol}$ http://pubchem.ncbi.nlm.nih.gov/compound/66320

This type of problem follows the pattern: $\text{mass of product}$$\rightarrow$$\text{moles of product}$$\rightarrow$$\text{moles of reactant}$$\rightarrow$$\text{mass of reactant}$.

Step 2: Convert mass of $\text{Li"_2"SO"_4}$ to moles $\text{LiSO"_4}$.
Divide the mass of lithium sulfate by its molar mass.

$250 \cancel{\text{g LiSO"_4xx(1"mol LiSO"_4)/(109.9446cancel"g LiSO"_4)="2.274 mol LiSO"_4}}$

I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.

Step 3: Convert moles of $\text{Li"_2"SO"_4}$ to moles of $\text{LiNO"_3}$.
Multiply moles of $\text{Li"_2"SO"_4}$ times the mole ratio between $\text{Li"_2"SO"_4}$ and $\text{LiNO"_3}$ from the balanced equation.

$2.274 \cancel{\text{mol Li"_2"SO"_4xx(4"mol LiNO"_3)/(2cancel"mol LiSO"_4)="4.548 mol LiNO"_3}}$

Step 4: Convert moles $\text{LiNO"_3}$ to mass $\text{LiNO"_3}$.
Multiply the moles $\text{LiNO"_3}$ times its molar mass.

$4.548 \cancel{\text{mol LiNO"_3xx(68.9459"g LiNO"_3)/(1cancel"mol LiNO"_3)="310 g LiNO"_3}}$ (rounded to two significant figures)

This problem can be worked out all at once as follows.

$250 \cancel{\text{g LiSO"_4xx(1cancel"mol LiSO"_4)/(109.9446cancel"g LiSO"_4)xx(4cancel"mol LiNO"_3)/(2cancel"mol LiSO"_4)xx(68.9459"g LiNO"_3)/(1cancel"mol LiNO"_3)="310 g LiNO"_3}}$ (rounded to two significant figures)

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