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 =

Answer 1

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

Answer:

0.242 molecules of Lithium Nitrate

Step-by-step 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

Answer 2

`1.46x10^(23)molecules \ LiNO_3`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`Li_2SO_4 + Pb(NO_3)_2 \rightarrow PbSO_4+2LiNO_3`

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:

`n_(LiNO_3)=13.3gLi_2SO_4*(1molLi_2SO_4)/(110gLi_2SO_4) *(2molLiNO_3)/(1molLi_2SO_4) =0.242molLiNO_3`

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

`0.242molLiNO_3*(6.022x10^(23)molecules\ LiNO_3)/(1mol)=1.46x10^(23)molecules \ LiNO_3`

Best regards.