L.3 Compounds that can be used to store hydrogen in vehicles are being actively sought. One reaction being studied for hydrogen storage is Li3N + 2 H2 ---> LiNH2 + 2 LiH. (a) What amount (in moles) of H2 is needed to react with 1.5 mg of Li3N? (b) Calculate the mass of Li3N that will produce 0.650 mol LiH.
How would I go about solving this question? I'm very lost, so I would appreciate the help! Thank you!
Reaction Stochiometry L.3 [ENDORSED]
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Re: Reaction Stochiometry L.3 [ENDORSED]
Hi!
For a, I would start by converting the 1.5 mg Li3N into grams of Li3N. After converting to grams, I would use the molar mass of Li3N to convert the grams of Li3N to moles of Li3N. Then I would this amount to convert to moles of H2 by using the "conversion factor" (2mol H2/1mol Li3N) to find the moles of H2 needed to react with 1.5 mg of Li3N.
Your set up should look like this:
1.5 mg=.0015 g Li3N
.0015g Li3N*(1 mol Li3N/34.83 grams)(2 mol H2/1 mol Li3N)
For b, I would use the "conversion factor" (1 mol Li3N/2 mol LiH) to convert 0.650 moles of LiH to moles of Li3N then use the molar mass of Li3N to find the mass of Li3N that will produce 0.650 LiH.
Your set up should look like this:
0.650 mol LiH *(1 mol Li3N/2 mol LiH)(34.83g / 1 mol Li3N)
Hope this helps!
For a, I would start by converting the 1.5 mg Li3N into grams of Li3N. After converting to grams, I would use the molar mass of Li3N to convert the grams of Li3N to moles of Li3N. Then I would this amount to convert to moles of H2 by using the "conversion factor" (2mol H2/1mol Li3N) to find the moles of H2 needed to react with 1.5 mg of Li3N.
Your set up should look like this:
1.5 mg=.0015 g Li3N
.0015g Li3N*(1 mol Li3N/34.83 grams)(2 mol H2/1 mol Li3N)
For b, I would use the "conversion factor" (1 mol Li3N/2 mol LiH) to convert 0.650 moles of LiH to moles of Li3N then use the molar mass of Li3N to find the mass of Li3N that will produce 0.650 LiH.
Your set up should look like this:
0.650 mol LiH *(1 mol Li3N/2 mol LiH)(34.83g / 1 mol Li3N)
Hope this helps!
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