Question

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Answer 1

2Al + 3H₂SO₄ = Al₂(SO₄)₃ + 3H₂

MAl = 27g/mol

MH₂SO₄ = 2AH+AS+4AO=2*1+32+4*16=98g/mol

a) 2*27g Al.......................3*98g H₂SO₄
2,7g Al...........................x=14,7g H₂SO₄ (md)

C=md/ms*100 => ms=md*100/C=14,7*100/4,9 = Calculeaza !!

b) n=m/M=2,7/27=0,1 moli Al

2 moli Al...........................3 moli H₂
0,1 moli Al........................x=3*0,1/2 = 0,15 moli H₂

c) 1 mol H₂.......................6,022*10^23 molecule
0,15 moli H₂......................x=Calculeaza !!

Answer 2

- ECUATIA REACTIEI
2mol      3mol          ..                         3mol
2Al + 3H2SO4 ------> Al2(SO4)3  + 3H2
2x27g....3x98g...................................3mol
2,7g.......m= .........................................n
m= 2,7x3x98/2x27=14,7gH2SO4   aceasta masa este md in solutia 4,9%
c/100= md/ms
4,9/100= 14,7/ms---->      calculeaza ms !!!

n,moliH2=2,7x3/2x27= 0,15molH2

1mol H2     are      N,A= 6,023X10²³molecule( numarul lui Avogadro)
0,15mol.......................N=