n=mA/MA
1=> n=mH2SO4/MH2SO4=9.8/(2+32+16*4)=0.1mol
2=>n=20/(64+12+16*3)=0.16mol
5=>n=31.5/(1+14+16*3)=0.5mol
6=>n=11/(12+16*2)=0.25mol
1. a/ Ta có: nH2SO4= \(\frac{m}{M}=\frac{9,8}{98}=0,1\) mol
b/ nCuSO4 = \(\frac{m}{M}=\frac{20}{124}\approx0,16mol\)
c/ nHNO3 = \(\frac{m}{M}=\frac{31,5}{63}=0,5mol\)
d/ nCO2 = \(\frac{m}{M}=\frac{11}{44}=0,25mol\)