Question

23) In the following equations, determine which reactant is the limiting reactant and which reactant is in excess. The amounts used are given below each reactant. Show evidence for your answers. a) KOH + HNO3 → KNO3 + H2O 16.0g 12.0g b)2 NaOH + H2SO4 → Na2SO4 + H2O 10.0 g 10.0g

Answer 1

Ericess reactanta koht 23 @) Hoog- Limiting reagent Ans! (6? Haso4 -Limiting reagent Excess reactant =Nooh 2 sol- Kho, + H₂O (4) Kott HNO3 63g as per Election 56g. 169 128 as pez problem needed First we need to calculate for 16g of kott, How much in grans HNO3 is required ? 63 g of Hilla → as per equation 56 g of kolt 16 g of Koh 9 A Loog 56 iz 18 g of HNO2 16 x63 56 1. x = less amount .. For 16g of kot needs 18g of HNO3, But given ing of Hnos only. Hence the one which is given than required is called called limiting reagent and the other one is Excess reagent . .. Limiting reagent = HNO3 Excess reagent - Kott

Na25oy & Hao 2 Naot + Ha so 4 2x40 g 988 as per Equation log og reacts as per Problem 989 H₂ Soy a as per equation Now 2x40g of Nooh log of naoh 2 x = we R$ 898 98 12.25 g of H₂Soy 8 2x40 anstead of 12.259 given only log of inson. need 18.259 To complete react with log of Naot & Hisou. . Limiting reagent: Hasou Excess reagent! Naoh. A 'gamo, moar mass Note ! - Sino substance NaoH 40 98 Idy soy 56 KOH 3 63 HNO3 다 How much product will get from the reaction is depends upon Limiting reagent.