Ile gramów tlenku sodu powstanie po spaleniu 46g sodu w tlenie?
mNa=23u
mNa2O=62u
4Na + O2 ---> 2Na2O
4*23g Na ----- 2*62g Na2O
46g Na --------- xg Na2O
x = 62g Na2O
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mNa=23u
mNa2O=62u
4Na + O2 ---> 2Na2O
4*23g Na ----- 2*62g Na2O
46g Na --------- xg Na2O
x = 62g Na2O