Question

4.4.52 g of the oxide of element X react with excess of H2 forming 2.56 g 8 points of pure element X and water. a) calculate mass of O in the oxide b) Calculate the quantity (mol) of H2O released in the reaction. C) What element is represented by the symbol X? d) How many liters of H2 (STP) are necessary to get 7.42 g of X from its oxide? Your answer

Answer 1

♡ Mass of Oxide = 4.52 grams Pure Element x = 2.56 grams Element pure mass of oxide- mass mass of ante Onygen (4-52-2.56) goms 1.56 grms H60 8:9 HO: Ratio of and Ralio of oilgo => a=2.205 grms The Quantity of so formed 8:9 = 1.9600 2.205 18 = 0,1225 mole m lo logo 8 8 He require for 2156 for 2156 gr of x :o.zurgoms My required for 7.42 gr of x = 0,245 x 7.42 0.71 grms 2214 x aial at STP, volume of agroftly = 22.4 litres Volume of Foll gr of h = 22,4 2 70952 litres ③ am not getting

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Rate of Reaction at which The Rate of reaction / speed of Reaction Reactants are Converted into change in concentration of per unit time Rate of reaction change products reactants [products Increase in product concentration in time. =