Question

Name CHEMISTRY 2A Study Assignment 11, Moles and Molar Mass Problems. Use dimensional analysis to solve the following problems. Pay attention to correct use of wants and correct use of significant figures in calculations 1) Convert 8.00 moles of aluminum to grams. A1= 26.98 8.00 mol A1 x 26,989 = 215,849 Im A 03 215.549 2) Convert 10.5 moles of aluminum chloride to grams. A=26.98 10.5 mol Alc1 X 62.429 = 655.53 cl=35.45 I mol All 655.59 3) Convert 0.250 moles of aluminum sulfate to grams. 4) Convert 2.70 grams of ammonia to moles. 5) Convert 0.2502 grams of calcium nitrate to moles. SA 11-1

Answer 1

We know 1 mole of atom/ moleculestion is weight is equal to the molecular mass . 1) 1 mole of Al notom's molecular mass = 26.98 g 8 moleu 26-98 = 1915.843) or in anothe way mole= to given may 1 Melecular May given mass = molex Molecular mars mass of Al = 8 x 26.98 = 215.849 1 mole of Alla molecules having mass = 133-342 Jo. 5 mole = 133.34 x10.5 = 1400.078 . I mole of Al(SO4)3 having mas = 342,15g 0.450 mile . . . . 342.15 X0.250 = 65.54 al ng of N ha having no. of mole = 1 mole ! . . . - *.72 2.70% . . . . -T0.159 mele 164.0 g of calcium nitrate having = 1 mole IX. 2502 1 , 164.0" 0.25023 . . . " .6015 mole 1 mole of Mg contain = 6.023x102 atom 3.85 moli = 6023X1023 X 3.80 A2.89 x122) atom 6.023 x 1023 atom of Copper contain a l mole . 1 atom . 6623 x 1923 34102" -0.418 10.498 mole 6.043 X 1023 molecules of com contain = 1 moleslis 1 molearles . 6.023X1023 9.25*10* . . -0.373 x 10-5 mole LX 3x10 IX 25 1018 - 10) 430.5 3 P Bes contain = 6.0 23X1023 molecules + 8 = 6.023 1023x192 430.5 192 à 5 - 2868 molecules -3 molculo 342.0 g man of sugar contain = 6.023x10²5 molecules To 1. = 6.023X10²3 40 X3420 = 60.23 moleculy = 60 molecules! 3.42 kg = 348