Question

N2(g) + 3F2 (g) 2.0 E 3 Lg) 6.718g of Nz 9.4219 of Fz N=14.00474 P=18.9984032 2) which is the limiting reactant? N₂ OR F2 How many grams of exuss reactant are ut over, assuming 100% yeild #if chemist recovered 1.519 g of NF₃- what is the percent yerld. 5) The reation type for- Alls) + Aud; laq) Micla laq) + Au(s) 4) The reaction type for- Lusoulag) + 2NaOH Lag) n(OH)2 (s) +Na₂504 (aq) I what is the net lonic equation for the reaction 3 CaCl2 (aq) + 2 Na3PO4 (aq) - Laz (PO4)2 (s) + NaCl lag) 5) when aquinous Solutions of ammonium carborvete and potassium sulfate are mixed, will a po precipate form? 4) for the reaction - which sons are spectator ons Zn(s) + CuSO4 Lag) ZnSO4 (ag) thuls). 10) An isothermal process is one in which the does not change (11) 12.16, gas occupied a volume of 4,938 L under expanded to 8.4142 1.595ata. - What is the 12) @ 18.3'c, gas occupied volume of 7-1852 under 2.457 am temperature of heated gas. heard + compressed : 1521 am What is new volume? + temp 81-7 1) pressure exerted by 5.492 a ot. R = 0.0820584 L atm/kmol what pressure? xerted by 5.492 g of tz in 2,893 2 @ 2037.8°

Answer 1

Ans following steps are used to find limiting Reagent O Calculate mole of each reactant Divide the mole of each reactants from their respective Stoichiometric coefficient 1 Reactant having least value will be limiting Reagent - 0.24 nele Mole Stoichiometric Coeft - N₂ + 3 F2 - 2 NF3 wt. of N2 = 6.718 jm Moie 0.24 0.36 Wt. Of Fc - 14.006 go is 0.24 0.36 = 0.12 Mole of N2= Mole of Fz - 140 . 0-36 mole The value of moleseth Ratio is least for F2, Hence in this Reaction Fz will act as limiting reagent. 28 3) Nz will act as excess reagente 3 Male Fz More F 0.36 mole Fz Reacts with Reacts with Reacts with I mox Nz 2 Mole Nz Ix0.36= 0.12 Mole Na Hence o.12 Mole of Me is used Remaining Moie of N2 = 0.24-0112 = 0.12 mole Im weight of No, which is left = 0.12 * 28 = 3-369m Ang