Question

Consider the balanced equation: 2 N2H4(g) + N2O4(g) + 3 N2(g) + 4 H2O(g) Complete the table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product formed. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made. mol N2H4 mol N204 mol N2 mol H2O 18 3.5 4.8 14.5 Copyright 2010

Answer 1

Ans: The balanced chemical equation of the reaction is 2N, 4419+ N204 (9) —+3N2 (9) + 442019) - © where no. of moles of Neha = 2 họ, P mºler PN, 0, 1 no. of moles of: N2 = 3 . . no. of moles of H2O = 4 . Now for solving the answer dets make the balanced equation in the more general form. Since no. of moles of N204=1 take it as a and moditieg other in terms of x' the ② becomes Q0N244 (9) + (0) N204 (9) —*@2) N2(9) +(429 M2 org) Foro Ist initial case) © x =1.. Now will troy to answer the different cases provided in the question... » Given: mol Na H4 .=2 jie. Q2). N244.= 2 e. x=1 > Using ☺ i.e. mol N204 = 1, mol Ng = 3, mol H₂O = 4 1 2. . QY Gven: mol N204 = 7 1. X.N204 = 7 i.e. x= 7 » mol N2 = 82 - 3X7 - 21 mol N, 74 -23 = 2X7 = 14 mol H20 - 42 -4X7=28 TV

3) Given mol H₂O = 18 48. H2O = 18, ie. x== 4.5 Then using egn . mol N2 H4 = 2X = 2X45 = 9 mol N₂O4 = x = 4.5 mol N2 - 32 = 884.5 = 13.5 147 Given: mol Na H4 = 3.5 22. N2H4 3.5 jie = 3: 5 1. Then using eqh ④ mol N204 = 2 = 1.75 mol N2 = 3 2 = 1.75X3 = 5.25 mol H2O = 4x.= 1.75*4 = 7.0. 1.45 2 2=4.8 - 5) Given: mol N,04 = 4.8 XN204 = 48 jie Now using egn mol. N2 H4 = 2x = 2x4.8 = 9.6 mol · N2 = B2 = 3*4.8 = 14.4 mol H₂O = 4x = 4x48 = 19.2 67 Given : mol N2 = 14.5 32. No = 14.5, x = 1425 7 48333 now using equ @ mol. N244 = 2x=2x 4.8333 = 9.6666 mol. N204 = x= 4.8333 mol H2O 4X19.3333 Scanned with CamScanner

Then our final table becomes : GEOGE (mol N2H4) (mot N204) 2 1 114 . 7. 19 . 14.5 3.5] [1.75 9.6 418) 9.6666 14.8333 (mol N2) 3 21 13:5] 5.25 14.4 114.5. Sol izo) 141 2 8] [18] 17.01 19.2 (19.3333 Note: [ 0 Given Anjwer Thank you Patauptils LX - Scanned with CamScanner