Opsta hemija

Submitted by Mery335 on Fri, 12/11/2015 - 21:55
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Izracunajte koliko CS2  i hlora je potrebno da se dobije 2,1345 g ClC4 i koliko S2Cl2 nastaje?

CS2+3Cl2=CCl4+S2Cl2

M(CS2)=76,2 g/mol  ; M(Cl2)=71,0 g/mol ; M(ClC4)=154,0 g/mol ; M(S2Cl2)=135,2 g/mol

Marija 96's picture

Marija 96 - Fri, 12/11/2015 - 23:00

m(ClC4)=2,1345g

m(S2Cl2)=?

n(ClC4)=2,1345g/83,5g/mol=0,0255mol

n(ClC4):n(CS2)=1:1

m(CS2)=0,0255mol*76=1,938g

n(Cl2):n(CS2)=3:1

n(Cl2)=n(CS2)*3=0,0765mol

m(Cl2)=0,0765mol*71=5,43g

n(S2Cl2)=0,0255mol

m(S2Cl2)=0,0255mol*135=3,44g

pklapa li se sa rješenjima?

Mery335's picture

Mery335 - Sat, 12/12/2015 - 06:58

Ne pisu rjesenja,ali mozes li mi pojasniti zasto si kad si racunala n(CCl4) podijelila sa 83,5 g/mol kad je M (CCl4)=154 g/mol

Marija 96's picture

Marija 96 - Sat, 12/12/2015 - 14:26

Pogresno sam napisala jedinjenje .Rdaila sam na brzinu. sad cu ponovo

m(CCl4)=2,1345g/154 =0,014mol

n(CCl4):n(Cl2)=1:3

n(Cl2)=3*0,014=0,042mol

m(Cl2)=0,042mol*71=2,982g

n(CCl4):n(CS2)=1:1

n(CS2)=0,014mol

m(CS2)=0,014mol*76g/mol=1,064g

n(S2Cl2):n(CCl4)=1:1

n(S2Cl2)=0,014mol

m(S2Cl2)=0,014mol*135g/mol=1,89g

 

Mery335's picture

Mery335 - Sat, 12/12/2015 - 15:18

Hvala ti ! :)