5g x(g)
NaCl + AgNO3 → NaNO3 + AgCl
Na 1*23 = 23
Cl 1*35.4 = 25.4
23 + 35.4 = 58.4g/mol
Ag 1*108 = 108
Cl 1*35.4 = 35.4
108 + 35.4 = 143.4g/mol
5g → x
58.4g/mol → 143.4g/mol
x = (5g)(134.4g/mol)
58.4g/mol x = 12.27g
2.- Que volumen de nitrogeno se obtiene de la reaccion de Bromo con Hidroxido de Amonio y se obtiene Bromuro de Amonio, Nitrogeno y agua, si se hace reaccionar con 250g de Bromo?
250g x(g)
3Br2 + 8NH4OH → 6NH4Br + N2 + 8H2O
6-2 ← Br → 1-6
8-1 ← N → 3-8
40-5← H → 6-40
8-1 ← O → 1-8
250g x(g)
3Br2 + 8NH4OH → 6NH4Br + N2 + 8H2O
3mol 8mol 6mol 1mol 8mol1mol = 22.4 Lts.
3(22.4Lts.) = 67.2 Lts.
n = m/PM = 250g =0.52mol
480g/mol
0.52mol → x
67.2 Lts. → 22.4 Lts.
x = (0.52mol)(22.4Lts.) x = 0.17mol
67.2 Lts.
V = n(22.4 Lts/mol)
V = 0.17mol (22.4 Lts./mol)
V = 3.80Lts.
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