You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. this reaction occurs 2ai(s)+ 3pb(no3)2(aq) -3pb(s)+ 2aino3la(aq) the theoretical yield of solid lead?

The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:  

 

Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

Moles of Aluminium:

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

For the given chemical reaction, the equation follows:

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = of lead metal.

Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

The answer is 65.3312

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2