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Liono4ka [1.6K]
3 years ago
8

For the reaction 6 Li + N2 → 2 Li3N , what is the maximum amount of Li3N (34.8297 g/mol) which could be formed from 14.18 mol (6

.941 g/mol) of Li and 16.37 mol of N2 (28.0134 g/mol)? Answer in units of mol.
Chemistry
1 answer:
Yuki888 [10]3 years ago
4 0
Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g
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Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.

Amount of acetaminophen initially taken = 350 mg

Amount of acetaminophen obtained after recovery =185 mg

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What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
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Missing in your question :

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Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

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3) equation 3 for Ka2 = 5.6 x 10^-11

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so, form equation 1& 2&3 we can get the overall equation:
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note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

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∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

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from the overall equation:

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so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
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