Another stoichiometry problem!?

Question

Pb(SO4)2 + 4 LiNO3 = Pb(NO3)4 + 2Li2(SO4)


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?


Show work please!



Correct answer will get you 10 points =)

Answer 1

1. go from grams of lithium sulfate to moles: 250g / 109.952 g = 2.27 mol
2. then multiply by the mole ratio (2 mols of lithium sulfate to 4 moles of lithium nitrate) = 4.547g lithium nitrate
3. then go from moles to grams 4.547 g X 68.951 g
4 the answer is 313.52 grams of lithium nitrate

hope that helps (:

Answer 2

250 g Li2SO4 x (1 mol Li2SO4/ 124 g Li2SO4) x (4 moles LiNO3/ 2 moles Li2SO4)x (69 g LiNO3/ 1 mol Li NO3)



= 278.23 g Li NO3

Answer 3

So you have 250g of Li2(SO4), you need to find the moles of that.

And you can find this in you periodic table.

Li= 6.9g/mol S=32 g/mol O=16 g/mol, now you add them all up....

(6.9)(2) + 32 + (16)(4) = 109.8 g/mol

So now to find the moles of Li2(SO4):

250g x 1mol/109.8g= 2.28 moles

Now you use the coefficients to find the moles of LiNO3..

2 moles of Li2(SO4) = 4 moles of LiNO3 so.....

2.28 moles Li2(SO4) x (4 moles of LiNO3/ 2 moles Li2(SO4))= 4.55 moles of LiNO3

Then find how many grams per mol LiNO3 has....so using the periodic table again
Li= 6.9 N=14 O=16
(6.9) + (14) + (16)(3) = 68.9g/mol

So now to find the grams of LiNO3 using the moles we found
4.55 moles x 68.9g/mol= 313.50 grams of LiNO3

And there's your answer.

Answer 4

FYI, we in many cases use the emblem "^" to intend "to the flexibility of". First, stability the equation 4Al + 3O2- ----> 2Al2O3 then you surely see that 3 moles of O2 react with 4 of aluminum to yield 2 moles of the trioxide. Now that's undemanding math to work out that 3*a million.40 4/4 = a million.08 mole of O2 will react with a million.40 4 moles of Al