Ile gramów tlenku fosforu (V) (P4O10) powstanie z utlenienia 0,1 mola tlenku fosforu (III)?
2P2O3 + 2O2 -->P4O10
mP4O10 = 4*31g/mol + 10*16g/mol = 124 + 160 = 284g/mol
2mol P2O3 --- 284g P4O10
0,1 mol P2O3 --- x
2x = 28,4
x = 14,2g
Odp: Powstanie 14,2grama P4O10.
Pozdrawiam ;)
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2P2O3 + 2O2 -->P4O10
mP4O10 = 4*31g/mol + 10*16g/mol = 124 + 160 = 284g/mol
2mol P2O3 --- 284g P4O10
0,1 mol P2O3 --- x
2x = 28,4
x = 14,2g
Odp: Powstanie 14,2grama P4O10.
Pozdrawiam ;)