Question

Cate grame de oxigen sunt continute in 2 moli de trioxid de azot;3 moli dioxid de carbon(CO2);4 moli oxid de magneziu.... Va rog :)

Answer 1

X g O 2 moli trioxid de carbon 3 moli COD 4 moli Mg O N2O3 MN2O3=2*14+3*16 = 28+48 = 76g 2 mOli...........x g n=m/Mai m=n*M=2*76=152 g 76 g N2O3.......48 g O 152 g.............. x X=152*48/76=96 g O Mco2=Ac+2*Ao=12+2*16=12+32=44 g m=n*M=3*44=132 g 44g COD. .......32g O 132 g........x X=132*32/44=96g O M MgO=AMg+AO=24+16=40g m=n*M=4*40=160g 40g MgO......16gO 160g..........x X=160*16/40=64gO