Question

Please do 5 and 6 only and can you show all your work of how to get the answer thanks.

on many wa ysel dumns are there in 91.2 grams of barium phosphate 3. How many grams of silver nitrate are needed to react with excess sodium phosphate to produce 42.7 grams of solid product? 4. Give the total molecular, total ionic, and net ionic reactions for each of the following: a. ammonium hydroxide plus hydrochloric acid b. calcium metal plus nitric acid d. sodium phosphate plus silver nitrate 5. Magnesium metal is exposed to a stream of nitrogen gas to form a binary compound containing only magnesium and nitrogen. From the following data, determine the empirical formula of the compound formed: Weight of crucible plus lid Weight of crucible plus lid plus magnesium Weight of crucible plus lid plus product 23.31 grams 25.92 grams 26.93 grams 6. A student was given an unknown metal chlorate and was asked to identify it by the professor. The student was told that the unknown could be cuprous chlorate, aluminum chlorate, or calcium chlorate. Upon heating the unknown metal chlorate, the student recorded the following data: Weight of crucible plus lid Weight of crucible, lid, sample Weight of crucible, lid, sample after heating 32.57 grams . 36.19 grams 34.50 grams When a student heated and decomposed 18.1 grams of sodium chlorate, 7.56 grams of gas were obtained. e percent error of this reaction? What is the maximum amount of solid product that can be produced by reacting 41.4 grams of potassium bonate with 59.8 grams of calcium nitrate? row an enerme diagram with the total energy of the reactants equaling 20 kJ, delta E equaling +28 kJ, 2

Answer 1

LELLULL p age- Answers lid 5. Mg(s) + N269) - → Mga Na (5). Magnesium Nitride. But, the formula of Magnesium nitride is to be found using mass %. .. Autor weight og Magnesium (sample) lot. gj ewyucible, fid, somple – wtig erucible, = 6(25.92 – 23.31) g = 2.619.. .. wt. of Mg = 2.61 g. Now, wt. of product = wt. of crucible, lid, product . ... - wt.g coucible, lid. = ( 26.93 – 23-31)9 = 3.62. :: wt of product (Magnesium Nitride) = 3.62.9 Lan, wt. o Nitrogen reacted a = .wt. Magnesium (Ng nitride - wt. Magnesium = (3.62 -2.61) g. = 1:01.9 .: wt og Mg= 2619. Mosť ay mass of Mg = 24.3 g/mol . moles J Mg present in Magnesium Nitidez given mass = 2.61 , Molar mass 24.3 = 0.1074 0.11 moles wt of N = 1.019, Molar mass of No.14 g/mol. i moles of N present in Magnesium Nitride = given mass su - = 100+ ration of moles a mig and N, 14 = 0.07 meter 0. Mg: N -0.11:0:07 = 1047 : 1 3 :2. Molar mass

.page-2 . Ratio g moles of Mg and N in compound Magnesium Nitride = 372. : Empirical foremula a is.Mg3 N2 06. wt. og sample - wt. of erucible, lid, sample - wtog erucible, lid = (36.19 – 32.57) 9 : = 3.62.97 wt.cf sample after heating = it. og eruciblé, lid, heated sample - Whoog erucible, lid = (34.50 – 32.579 = 2.93g. .: The oxygen gas release (wt.) = 3.62 -1.93 1:69.9 cerca elO3)2 → cuela + 302 on, 1 mole. Cu( C103)2 gives 3 mole oz.19). moles gi 02. sieleased = moles of 02- Heleased given mass, = 10699 = 0.05 Molar mass 7 329/mol mole.

1.5 = 4.85 g hence, Cuptocs chlorake, even . → Cuci + 1.502 (pase-3 1 mole of eucloz gives 1.5 Oz. ..;0.05 mole oz given by 1 X0.05 = 0.033 mole Cu C103. · Now, mass of 0.0.33 mole Cu C103 = 0.033 x Molar mass = 0.033 x 147 Hence, not matening the wti oj sample, hence, sample is not Cu C103 ( cup rows chiosale): aluminium chlorată Al (403), → Alclz + 4.502. 4.5 miele 02 given by 1 mole. Al(C103)3. ::: 0:05 mole 02 given by 1x0505= 0.011 mole. Al(103)3. :: mass of 0.011 mole Al (1103)3 = 0.011* Molay mass. = 277.5 x 0.011. 3-0 525. Hence, not matching the wtor sample, hence, sample is not Al (1103)3. calcium chlorale, la (1103)2 - Atels Call2+ 302 3 mole 02 giver by I mole caſc103)2. .:0.05 mole 02 given by x0.05 -0.016 mol cale103) mass of 0.016 mol calc103 )2 = 00026 X Molay mass = 0.016* 207. 3 3.312.9 The value is close to the wife of sample. Hence, the unknown sample is calcium Chlorale Ca (6103)2