Hi! So I\'ve done part A-E on question 10 & section A on question 11. I\'d like

Question

Hi! So I've done part A-E on question 10 & section A on question 11. I'd like to know if I did them correctly, but more importantly I am struggling with part 10 F and 11 B and C . Teacher seems to make things more difficult to understand than they need to be, so if you have a better way of explaining I would greatly appreciate it! He writes the "textbook" too so I don't have much to refer to. Thank you!.

n definc cespomei to tdut of ow many grams af co can be obtained from the combustion of 10 grams r ce on's expact lo espoct to pass Cheniary less you coan tande thoe kinds f problors uestion 10. a) Use the bottom number on the element's periodic table box to the grams (Q) in 1 mole of each element below. I mole Cas 5) i mole C 1 mole S g Ces This se your answers to question 10a to fil n the Ca ond S blanks above hemist how much Ca and S to weigh our c) Add the component masses to get the moss of one male of Cas. Put r ass in the g Ca5 blank in question 10 b. g Cas ) Use the g values in question 10 b to fag in the conversion factor below e Each 5 atom has to hit a Ca atam to get the 2 electrons to fill its hell Use the conversion factor in question 10.d on 20 g 5 to see how m il be needed to react it all

Explanation / Answer

! Atomic mass of Calcium =40 and atomic masss of sulfur =32

the reaction between Calcium and sulfur can be represented as Ca+S--->CaS

1 mole of Calcium reacts with 1 mole of sulfur to form 1 mole of CaS

40.08 gms of Calcium r reactas with 32.07gms of sulfur to form 72.15 gms of CaS

32.07 gms of sulfur requires 40 gms of Calcium

20 gm of calcium requires 20*40/32.07=24.94 gms So this is correct

b) 40.08 gms of Calcium reacts with adequate sulfur to produce 72.15 gm of CaS

80 gms of Calcium reacts with adequate sulfur to produce 80*72.15/40.08 gm of Cas=144.012 gms of CaS

2. 1 mole of N2= 28 gm, 1mole of H2 =2 gms, Ammonia is NH3 and 1 mole NH3= 17 gm (14+3=17)

the reaction is N2+3H2---->2NH3

28 gm of N2 + 6 gms of H2     gives 34 gm of NH3

42 gm of N2 requires 42*6/28 gm of H2=9 gm of H2

6 gms of hydrogen with sufficient nitrogen produces 34 gm of NH3

42 gm of hydrogen with sufficient nitrogen produces 42*34/6 =238 gm of NH3

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