Question

Question 4 Given the decomposition reaction of KCIO3(s), 2 KCIO3(s) 2 KCl(s) + 3 O2(g) If 46.5 L of oxygen gas (O2) are produced at 208 °C and 760 torr, answer all three parts below. 1) The number of moles of O2(g) produced is Select 1 (Select) 2) The number of moles of KCIO3(s) required 895 mol nt of O2(g) as 2070 mol stated in the problem is [Select 1 2.72 mol 1.18 mol 2) The grams of KCIO3(s) required to produced the amount of O2(g) as stated in the problem is Select 1

Answer 1

11 a) 2 Kcles 67 - 2Kcle, +30, 19/ calculation of number of moles of a produced of pu= nRT C P = 760 tore & latm? non habet po 1= 2089 +273 k= 4812 tos 46.5L R=0.0821 L. arm. Ilmoit 22_21_446.84, n I alm X 46.52 0.0821 L.atm. Kimoil x 481k no = 1.177 mole by calculation of mass of kelog required. According to the given balanced equation 2 moles of kelog produces, 3 moles of O how many molen of kelog produces 1.177 moles of a the - 103 1.177 mole x 2 mole z mole kolo = 0.7846 mole --Mass of keloz i molecular mass of ķeloz n keloja

Mass of kelog = ncclo x molecular mass - mass of kul@z = 0.7846 mole x 122.59Imol Mass of kclos_= 96-41359 (Note: molecular mass of kelo;= 122.5g|mad)