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

Answer:

$1.46x10^{23}molecules \ LiNO_3$

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:

$n_{LiNO_3}=13.3gLi_2SO_4*\frac{1molLi_2SO_4}{110gLi_2SO_4} *\frac{2molLiNO_3}{1molLi_2SO_4} =0.242molLiNO_3$

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

$0.242molLiNO_3*\frac{6.022x10^{23}molecules\ LiNO_3}{1mol}=1.46x10^{23}molecules \ LiNO_3$

Best regards.

Answer 2

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

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