Question

5. Potassium chlorate reacts with hydrochloric acid to produce chlorine gas. What mass of KCIO3 is needed to make 4.50 L of chlorine at 2.35 atm and 37.0°C? KCIO3(g) + 6HCl(aq) → KCl(aq) + 3Cl2(g) + 3H2O(1)

Answer 1

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1 Lietumor! Kelo, (g) + 6 Hel can) - Kce (aq) + 3Cl₂ (9) + 3 H₂0 (4) & Post Calculate the no. of moles of chlorine gas using ideal gas equation. TPV = nRT P = pressure = 2.35 atm; n = no. of moles = ?? V = volume = 4.50L R = Gas constant = 0.0821 L. atm. motto T= Temperative = 37.0°C+273 = 310k → n= PV = (2.35 ath (4.504) 10.575 mol (0.0821x.atm. mot 12 (210 k). 25-451 n = 0.415 mol From equation, 1 mot klo produce & not cla (1: 3 mol ralis) so - mot kuo produce 0.415 mol Cl, & Imol x 0.415 mas = 0. 1385 mot of kuo; 3 mol # Now, calculate the mass of kulo no. of moles = mass 6 To a mass = no. of ma x molar mass molar mass (g/mol) on mass = 0.1385 mol X 122.55 g/mol = 16.973 g of kao is required