In the reaction between Lithium Sulfate and an excess of Lead (II) Nitrate, how many molecules of Lithium Nitrate can be expected as a product if 13.3 grams of Lithium Sulfate are used? Show your work. Li2SO4 + Pb(NO3)2 =

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Thus, since lead (II) nitrate is in excess, we can directly compute the moles of lithium nitrate by applying the 1:2 molar ratio between them in the chemical reaction as well as the molar mass of Lithium Sulfate that is 110 g/mol for the stoichiometric shown below factor:

Finally, by using the Avogadro's number we are able to compute the molecules:

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Complete question:

In the reaction between Lithium Sulfate and an excess of Lead (II) Nitrate, how many molecules of Lithium Nitrate can be expected as a product if 13.3 grams of Lithium Sulfate are used? Show your work.

Li₂SO4 + Pb(NO3)₂ = LiNO₃ + PbSO₄

0.242 molecules of Lithium Nitrate

Explanation:

Given:

Li₂SO₄ + Pb(NO3)₂ =  LiNO₃ + PbSO₄

A balanced form of the equation:

Li₂SO₄ + Pb(NO3)₂ =  2LiNO₃ + PbSO₄

When  13.3 grams of Lithium Sulfate (Li₂SO4 ) are used, the number of molecules of Lithium Nitrate (LiNO₃) that can be expected as a product is calculated as;

molecular mass of  Li₂SO₄  = (7 x 2) + (32) + (16 x 4)

                                             = 14 + 32 + 64

                                             = 110 g

110 g of  Lithium Sulfate > 2 molecules of Lithium Nitrate

13.3 g of  Lithium Sulfate > ? molecules of Lithium Nitrate

= (2 x 13.3) / 110

= 0.242 molecules of Lithium Nitrate

Therefore, 13.3 grams of Lithium Sulfate will produce 0.242 molecules of Lithium Nitrate