Question

How many grams of the first reactant in each of the following chemical equations would be needed to produce 10.8 g of N, gas? (a) 4NH3 + 30, 2N2 + 6H,0 g of NH3 (6) (NH4)2Cr20, – N2 + 4H,0 + Cr203 g of (NH4)2Cr2O7 (c) N,H4 + 2H02N2 + 4H20 g of N2H4 (d) 2NH3 – N2 + 3H2 g of NH3

Answer 1

WALL molar mass of N2 .=28001 I mot g/mol No of motes in 18²8g N2= 10 8 g = 0.39 mot. 1 28-01 g/mol * 4 NH3 +382 - 2N2 + 6H₂O from the reaction. 2 motes of N₂ get from = 4 moles of NH3 .: 8-34 meters of Nz get from= 4 10:39 moles of NH3 R = 0.78 meter of NH₃ molar mass of NH₃ = 17 g (mol .) 0.48 mole” = 17 * 0°78 513262, Ms M43 ® (NH, , Ch.0, 2 + 4 1.0 +21.02 here, y mates of Mz frem I motes of (NHala Chez Oz :: 039 moles of Nz forem 0' 39 modes of (NH4), CH2 Oz Molar mass off (NH4), Chzla = 252 glmal "1 039 motos of (NHW), CA2 D.a = 0°39 x 252 =9828 g *

. N₂ Hu + 2 H₂O₂ N₂ + 4760 here, ' I man 1 modes of Nz from I moles of Natta - 0.39 moes of Nz from 0:39 moles of Nattu Molar mass of Natty = 32-04 g/mol :: 0:39 mote of Natlu = 32:04 X0:39 - 12:49 g y of N. Ha ♡ 2NH3 - Net 3 Hz are 1 mote Nz vrum 2 mole NH3 10:39 mole N2 = 2X 0.39 moles NH₂ = 0.78 meter Molar mass of NH3 = 17 g/mole : 0.78 moles = 13:26 g) of NH3