JEE Main 2019 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2019 (Online) 10th April Evening Slot
Question
The minimum amount of O2(g) consumed per gram of reactant is for the reaction :
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
Choose an option
Show full solutionCorrect option: A
Correct answer
A4Fe(s) + 3O2(g) 2Fe2O3(s)
Step-by-step explanation
(a) 4Fe(s) + 3O2(g) 2Fe2O3(s)
Moles of O2 = moles = moles = g of O2 = 0.43 g of O2
(b) P4(s) + 5O2(g) P4O10(s)
Moles of O2 = 5 moles = moles = g of O2 = 1.3 g of O2
(c) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Moles of O2 = 5 moles = moles = g of O2 = 3.6 g of O2
(d) 2Mg(s) + O2(g) 2MgO(s)
Moles of O2 = moles = moles = g of O2 = 0.66 g of O2
Moles of O2 = moles = moles = g of O2 = 0.43 g of O2
(b) P4(s) + 5O2(g) P4O10(s)
Moles of O2 = 5 moles = moles = g of O2 = 1.3 g of O2
(c) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Moles of O2 = 5 moles = moles = g of O2 = 3.6 g of O2
(d) 2Mg(s) + O2(g) 2MgO(s)
Moles of O2 = moles = moles = g of O2 = 0.66 g of O2
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