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N2O5+MgO--->Mg(NO3)2
16g xg
40u---148u
16g---xg
x=16g*148u/40u
x=59,2 g
Masa molowa MgO = 24+16
= 40 g/mol
Masa molowa Mg(NO3)2 = 24 + 14*2+16*6
= 148 g/mol
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Układam proporcję:
148 g Mg(NO3)2 ------------ 40 g MgO
x g Mg(NO3)2 ---------------- 16 g MgO
x=59,2g azotanu V magnezu