See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let no. of moles of CH4 present = n1 mol. Let no. of moles of C2H2 present = n2 mol. CH4 + 2O2  CO2 + 2H2O n1 n1 C2H2 + 2 5 O2 CO2 + H2O n2 2n2 Total no. of moles at initial = n1 + n2 Total no. of moles at final = n1 + 2n2 At constant V & T, 2 1 P P = final at moles of . No initial at moles of . No 69 63 = 2 1 2 1 n 2 n n n   23 21 = 2 1 2 1 n 2 n n n    1 2 n n = 19 2  1 2 1 n n n  19 21  2 1 1 n n n  21 19