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CHE1501 ASSIGNMENT 2 SEM1 2021(Latest tutorial letter)
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This document contains CHE1501 Assignment 1 Solutions(Latest tutorial letter). All workings are shown clearly and explanations are provided.
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CHE1501ASSIGNMENT 2 2021
QUESTION 1
𝐶2 𝐻6 + 5𝑂2 → 3𝐻2 𝑂 + 2𝐶𝑂2 𝐶𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
ANSWER:[4]
QUESTION 2
𝐵𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛:
2𝐴𝑔𝑁𝑂3 + 𝐶𝑎𝐶𝑙2 → 2𝐴𝑔𝐶𝑙 + 𝐶𝑎(𝑁𝑂3 )2
ANSWER:[4]
QUESTION 3
𝐴𝑐𝑖𝑑 𝑏𝑎𝑠𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛:
𝐻𝑦𝑑𝑟𝑜𝑐ℎ𝑙𝑜𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 + 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑥𝑖𝑑𝑒 → 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒 + 𝑤𝑎𝑡𝑒𝑟
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 = 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒 𝑎𝑛𝑑 𝑤𝑎𝑡𝑒𝑟
ANSWER:[2]
QUESTION 4
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝑛𝑒𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑, 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑟 𝑖𝑜𝑛
𝑎𝑛𝑑 𝑎𝑙𝑠𝑜:
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝑠𝑢𝑚 𝑜𝑓 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑚𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑝𝑖𝑐𝑖𝑒𝑠
𝐿𝑒𝑡: 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻 = 𝑥
,𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑎 = +2 (𝐺𝑟𝑜𝑢𝑝 2 𝑒𝑙𝑒𝑚𝑒𝑛𝑡)
𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑎 + 2(𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻) = 𝑛𝑒𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
2 + 2(𝑥) = 0
2𝑥 = −2
𝑥 = −1
𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻 = −1
ANSWER:[2]
QUESTION 5
∗ 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻 𝑖𝑠 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 + 1 𝑒𝑥𝑐𝑒𝑝𝑡 𝑖𝑛 𝑚𝑒𝑡𝑎𝑙 ℎ𝑦𝑑𝑟𝑖𝑑𝑒𝑠
∗ 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑂 𝑖𝑠 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 − 2 𝑒𝑥𝑐𝑒𝑝𝑡𝑠 𝑖𝑛 𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒𝑠
𝐿𝑒𝑡: 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑛 = 𝑥
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑁𝐻4 + = +1
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑂 = −2
2(𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑁𝐻4 + ) + 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑛 + 4(𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑂) = 0
2(+1) + 𝑥 + 4(−2) = 0
𝑥 = −2 + 8 = +6
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑛 = +6
ANSWER:[4]
QUESTION 6
𝐿𝑒𝑡: 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃 = 𝑥
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐾 = +1 (𝐺𝑟𝑜𝑢𝑝 1 𝑒𝑙𝑒𝑚𝑒𝑛𝑡)
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻 = +1
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑂 = −2
, 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐾 + 2(𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐻) + 𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃
+4(𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑂) = 0
1 + 2(+1) + 𝑥 + 4(−2) = 0
1+2+𝑥−8=0
𝑥−5=0
𝑥 = +5
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃 = +5
ANSWER:[4]
QUESTION 7
∗ 𝑂𝑥𝑖𝑑𝑖𝑧𝑖𝑛𝑔 𝑎𝑔𝑒𝑛𝑡 𝑎𝑙𝑤𝑎𝑦𝑠 𝑔𝑎𝑖𝑛𝑠 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑎𝑛𝑑 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑒𝑑
ANSWER:[4]
QUESTION 8
ANSWER:[4]
QUESTION 9
𝐿𝑒𝑡: 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑛 𝑖𝑛 𝐾𝑀𝑛𝑂4 = 𝑥
𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐾 = +1
𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑂 = −2
1 + 𝑥 + 4(−2) = 0
𝑥+1−8=0
𝑥 = +7
𝐿𝑒𝑡: 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑛 𝑖𝑛 𝑀𝑛𝑆𝑂4 = 𝑦
𝑦 + (−2) = 0
𝑦 = +2
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