The experiment is the reduction of a nitro group to an amine I've attached my work, but I'm not sure if it's correct. Can you please double check that my reduction equation is correct and then that my calculations are correct (especially the limiting reactant and theoretical yield) This is the reduction of m-nitroacetophenone and granular tin to m-aminoacetophenone We are using 200 mg of m-nitroacetophenone and 400 mg of granular tin. This will be done with 4mL of 6M hydrochloric acid and 30% sodium hydroxide.

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Ohem lab question The experiment is the reduction of a nitro group to an amine I've attached my work, but I'm not sure if it's correct. Can you please double check that my reduction equation is correct and then that my calculations are correct (especially the limiting reactant and theoretical yield) This is the reduction of m-nitroacetophenone and granular tin to m-aminoacetophenone We are using 200 mg of m-nitroacetophenone and 400 mg of granular tin. This will be done with 4mL of 6M hydrochloric acid and 30% sodium hydroxide.
Balanced Redox Equation?
4+
12H² + 2 Cg Hy NO₂ + 3 Sn 2 Cg Hq NO + 3 Sn² + 4 H ₂0
Given: 200 mg of m-nitroacetophenone (Cg H7 NO3)
of
tin (Sn)
400 mg
4mL
granular
of 6 M hydrochloric acid
Mass of m-nitroacetophenone
200 mg 19
Moles of m-nitroacetophenone
moles of
mass of granular tin
0.2g/ I mol = 0.0012 mol C8H7N03
165.159
400mg 1g
1000mg
granular
1000mg
Limiting Reactant
tin
=
0.4g|Imol
118.719
Theoretical Yield
0.0012 mol CxHyN03
=
:
0.29 C8H7 NO3
0.4g Sn
0.0034 mol Sn
0.0012 mol CH₂ No3 mol CgHqNO
a mol C8 H7 NO3
2 mol C8HqNU
2 mol (8H7 NO3
0.0034 mol Sn 2 mol C& HqNO
3 mol Sn
*m-nitroacetophenone is the limiting reactant
= 0.0012 mol C8Hq NO
=
0.0023 mol C8Hq NO
135.179 C8HqN0-
Imol C8HqNo
0.162 g
C8HqNO
Transcribed Image Text:Balanced Redox Equation? 4+ 12H² + 2 Cg Hy NO₂ + 3 Sn 2 Cg Hq NO + 3 Sn² + 4 H ₂0 Given: 200 mg of m-nitroacetophenone (Cg H7 NO3) of tin (Sn) 400 mg 4mL granular of 6 M hydrochloric acid Mass of m-nitroacetophenone 200 mg 19 Moles of m-nitroacetophenone moles of mass of granular tin 0.2g/ I mol = 0.0012 mol C8H7N03 165.159 400mg 1g 1000mg granular 1000mg Limiting Reactant tin = 0.4g|Imol 118.719 Theoretical Yield 0.0012 mol CxHyN03 = : 0.29 C8H7 NO3 0.4g Sn 0.0034 mol Sn 0.0012 mol CH₂ No3 mol CgHqNO a mol C8 H7 NO3 2 mol C8HqNU 2 mol (8H7 NO3 0.0034 mol Sn 2 mol C& HqNO 3 mol Sn *m-nitroacetophenone is the limiting reactant = 0.0012 mol C8Hq NO = 0.0023 mol C8Hq NO 135.179 C8HqN0- Imol C8HqNo 0.162 g C8HqNO
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