hope it will help you
.
.
.
hope it will helps
When 70. g of Li3N(s) (molar mass 35 g/mol) reacts with excess H2(g), 8.0 g of LiH(s) is produced. The percent yield is closest
1 answer:
The percent yield is closest to 17%
<h3>How to calculate the percentage yield</h3>The reaction between Li3N(s) and LiH(s) is given as:
Calculate the moles of Li3N
Moles = Mass/Molar mass
Moles = 70/35
Moles = 2 moles
From the reaction you can see that 1 mole of Li3N produce 3 moles of LiH, hence the moles of LiH is 6 moles
Determine the mass of LiH (theoretical yield)
Mass of LiH = 6 * 7.95
Mass of LiH = 47.7g
%yield = actual/theoretical * 100
%yield = 8/47.7 * 100
%yiels = 16.77%
Hence the percent yield is closest to 17%
Learn more on percent yield here: brainly.com/question/25996347
You might be interested in
Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.
