Question

40g fosfor se combina cu 64g oxigen. Este reactia totala? Daca nu calculati substanta in exces. Ce cantitate P2O5 se obtine? URGENT

Answer 1

MP=31g/mol

MO2=32g/mol

n=m/M=40/31=1,29 moli P

n=m/M=64/32=2 moli O2

4P + 5O2 => 2P2O5

4 moli P..................5*32g O2
1,29 moli P.............x g O2

x=5*32*1,29/4=51,6g O2

m exces = 64-51,6=12,4g O2

4 moli P.......................2 moli P2O5
1,29 moli P..................x moli P2O5

x=2*1,29/4=0,645 moli P2O5

MP2O5=2AP+5AO=2*31+5*16=142g/mol

n=m/M=>m=n*M=0,645*142= Calculeaza !!

Answer 2

ecuatia reactiei este
4mol      5mol          2mol--------transform in grame
4P  +    5O2-------> 2P2O5
4x31    5x32............2(  2x31gP+5x16gO)
40.............X....................Y
X= 40x5x32/4x31  g=51,6gO2; cum s-a dat m,O2=64, rezulta ca O2 este in exces--> excesO2= 64-51,6=...........................

y=40x2 x142/4x31 g=......................calculeaza gP2O5

VALORILE MASELOR ATOMICE SUNT EXTRASE DIN S.P.