Question

16. When 3.5 g of Fe combines with elemental oxygen, it forms 5.0 grams of a binary compound Fe.O, - made of Fe and O atoms only. Find its empirical formula (3 pts): In the 5.0 gram quantity of the compound, the mass of the Fe atoms is In the 5.0 gram quantity of the compound, the number of moles of the Fe atoms is: (Fe) (Fe) x mol(Fe) (Fe) In the 5.0 gram quantity of the compound, the mass of the O atoms is In the 5.0 gram quantity of the compound, the number of moles of the O atoms is: (O) g(0) - mol(0) (0) mol(Fe) The empirical formula (EF) is: - » Fe o mol(0) 17. Convert the word into formula equations, add states, balance and classify (4 pts): Solid copper(II) chloride with elemental hydrogen solid copper and hydrogen chloride gas: classify: Solid aluminum with (aq) copper(II) nitrate (aq) aluminum nitrate and solid copper classify: Classify these and all other reactions below by changes in composition (COM, DEC, and/or SDP, or DDP), and as redox (R/O) and combustion (CMB), if applicable. Page 3 of 6

Answer 1

16) 3.58 fe 59 binary compound fex Oy mass of product = mass of Fe + mass of oxygen mass of oxygen= 6-3.5) 8 = 1-58 moles of fe= 3.58 55.8453/mol = 0.062 mol Molar mass of moles of o = fe= 55.8458/mol 1:58 168/mol = 0.093 mos Then we take the molar ratio of feand o Moles of re: Moles of o 0.062 : 0.093 Divided by smallest one 0.062 10.093 0.062 . 0.09363 1 :1.47 mallst whole number ratio that Smal nts multiply both by 2 the t etween cs Scanneffe, 01.47)2 = feroz CamScanner.

In the 5.0 from auantity of the compound, the mass of the featomis 3-5 8 In the 5.0 gram alantity of this compound, the moles of fe atoms is 3.58 Cre) x imel feia - = 0.062 melle 55.845 9 Fe. In the 5.0 gram of the compound, the mass of O atoms is = 1:58 The mores of o atoms is. -590) x. i mollo) 2 = 0.093 molo) 1680) The Efis 0.062 mol(fe). 30.093 mol ☺ = 0.666 foco fe 2 02 & es Scanned with Home Camscanner W WW