CHE1503 ASSIGNMENT 1 2023

CHE1503
ASSIGNMENT 1
2023

, QUESTION 1




Solution:



𝑆𝑂3 (𝑔) + 𝐻2 𝑂 (𝑙) → 𝐻2 𝑆𝑂4 (𝑎𝑞)



𝑚𝑎𝑠𝑠 (𝑆𝑂3 ) 𝑢𝑠𝑒𝑑 = 40.1 𝑔



𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑆𝑂3 ) = 32.06 + 3(16)
= 80.06 𝑔⁄𝑚𝑜𝑙


𝑚𝑎𝑠𝑠(𝑆𝑂3 )
𝑚𝑜𝑙𝑒𝑠 (𝑆𝑂3 ) 𝑢𝑠𝑒𝑑 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠(𝑆𝑂3 )
40.1
𝑚𝑜𝑙𝑒𝑠 (𝑆𝑂3 ) 𝑢𝑠𝑒𝑑 =
80.06
𝑚𝑜𝑙𝑒𝑠 (𝑆𝑂3 ) 𝑢𝑠𝑒𝑑 = 0.5 𝑚𝑜𝑙




𝑣𝑜𝑙𝑢𝑚𝑒 (𝐻2 𝑂 ) = 10𝑚𝐿 = 10 𝑐𝑚3



𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝐻2 𝑂 = 1 𝑔⁄𝑐𝑚3



𝑚𝑎𝑠𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝐻2 𝑂 × 𝑣𝑜𝑙𝑢𝑚𝑒 (𝐻2 𝑂 )
𝑚𝑎𝑠𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 = 1 × 10
𝑚𝑎𝑠𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 = 10 𝑔

, 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝐻2 𝑂 ) = 2(1,008) + 16
= 18.016 𝑔⁄𝑚𝑜𝑙


𝑚𝑎𝑠𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑
𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝐻2 𝑂 )
10
𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 =
18.016
𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 = 0.555 𝑚𝑜𝑙




1 𝑚𝑜𝑙 𝐻2 𝑂
𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 0.5 𝑚𝑜𝑙 𝑆𝑂3 ×
1 𝑚𝑜𝑙 𝑆𝑂3
𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 0.5 𝑚𝑜𝑙 𝐻2 𝑂



∴ 𝑆𝑖𝑛𝑐𝑒 𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 < 𝑚𝑜𝑙𝑒𝑠 (𝐻2 𝑂 ) 𝑢𝑠𝑒𝑑 , 𝑆𝑂3 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑟𝑒𝑎𝑔𝑒𝑛𝑡.



𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑎𝑙𝑙 𝑆𝑂3 𝑟𝑒𝑎𝑐𝑡𝑒𝑑:


1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = 0.5 𝑚𝑜𝑙 𝑆𝑂3 ×
1 𝑚𝑜𝑙 𝑆𝑂3
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = 0.5 𝑚𝑜𝑙 𝐻2 𝑆𝑂4


4.9 𝑚𝑜𝑙 𝐻2 𝑆𝑂4 1𝐿
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = × × 20.1 𝑚𝐿
𝐿 1000 𝑚𝐿
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = 0.09849 𝑚𝑜𝑙 𝐻2 𝑆𝑂4


𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
% 𝑦𝑖𝑒𝑙𝑑 = × 100%
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
0.09849
% 𝑦𝑖𝑒𝑙𝑑 = × 100%
0.5
% 𝑦𝑖𝑒𝑙𝑑 = 19.7%