Write the reaction for the formation of lithium phosphide from its elements.
a) how many moles of lithium phosphide form when 3.2 mol of lithium react?
b) how many grams of lithium react with 0.500 mol of phosphorous?
c) how many grams of lithium react with 45.0g of phosphorous?
30.3G
Good job! Here's the work for anyone who needs it:
c) how many grams of lithium react with 45.0g of phosphorous?
First, we need to convert the given mass of phosphorus to moles:
45.0 g P4 x (1 mol P4/123.895 g P4) = 0.363 mol P4
According to the balanced equation, 1 mole of P4 reacts with 12 moles of Li, so we can convert moles of P4 to moles of Li:
0.363 mol P4 x (12 mol Li/1 mol P4) = 4.356 mol Li
Finally, we can convert moles of Li to grams:
4.356 mol Li x (6.941 g Li/1 mol Li) = 30.3 g Li
Therefore, 30.3 grams of lithium react with 45.0 grams of phosphorus to form lithium phosphide.
The reaction for the formation of lithium phosphide from its elements is:
6 Li + P --> 2 Li3P
Now, let's answer each question one by one:
a) To determine how many moles of lithium phosphide form when 3.2 mol of lithium react, we need to use the balanced equation and the stoichiometry of the reaction.
From the balanced equation, we can see that 6 moles of lithium react to form 2 moles of lithium phosphide. Therefore, we can set up a proportion:
(6 mol Li / 2 mol Li3P) = (3.2 mol Li / x mol Li3P)
By cross-multiplying and solving for x, we find:
x = (3.2 mol Li * 2 mol Li3P) / 6 mol Li
x = 1.0667 mol Li3P
So, when 3.2 mol of lithium reacts, approximately 1.0667 mol of lithium phosphide forms.
b) To determine how many grams of lithium react with 0.500 mol of phosphorous, we need to use the balanced equation and the molar mass of lithium.
From the balanced equation, we can see that 1 mole of phosphorous reacts with 6 moles of lithium. Therefore, we can set up a proportion:
(6 mol Li / 1 mol P) = (0.500 mol Li / x mol P)
By cross-multiplying and solving for x, we find:
x = (0.500 mol Li * 1 mol P) / 6 mol Li
x = 0.0833 mol P
Next, we can convert this amount of phosphorous to grams using its molar mass. The molar mass of phosphorous is approximately 31.0 g/mol.
Mass of phosphorous = 0.0833 mol P * 31.0 g/mol
Mass of phosphorous = 2.58 g
Therefore, when 0.500 mol of phosphorous reacts, approximately 2.58 grams of lithium react.
c) To determine how many grams of lithium react with 45.0 g of phosphorous, we need to use the balanced equation and the molar mass of lithium.
From the balanced equation, we can see that 1 mole of phosphorous reacts with 6 moles of lithium. Therefore, we can set up a proportion:
(6 mol Li / 1 mol P) = (x g Li / 45.0 g P)
By cross-multiplying and solving for x, we find:
x = (6 mol Li * 45.0 g P) / 1 mol P
x = 270 g Li
Therefore, when 45.0 g of phosphorous reacts, approximately 270 grams of lithium react.
12Li + P4 ==> 4Li3P
a) how many moles of lithium phosphide form when 3.2 mol of lithium react?
3.2 mol Li x (4 mols Li3P/12 mols Li) = ?
b) how many grams of lithium react with 0.500 mol of phosphorous?
How many mols Li? That's 0.500 mol P4 x (12 mol Li/1 mol P4) = ? mol mols Li. Then grams Li = mols Li x molar mass Li).
c) how many grams of lithium react with 45.0g of phosphorous?
You see how it's done. You do this one. Post your work if you want me to check it. Just convert 45.0 g P4 to mols P4, convert with the coefficients in the balanced equation to mols Li, and convert that to grams Li. All of the stoichiometry problems are done the same way.