Gọi nNO=x mol, nN2O=y mol
=> x+ y= 0,25 mol
Ta có PT electron:
nMg*2 =nNO*3+ nN2O*8
<=>\(\frac{15}{24}\cdot2=x\cdot3+y\cdot8\)
<=>3x+8y=1,25 mol
=> x=0,15, y=0,1 mol
=> VNO=3,36l, VN2O=2,24l
=> %VNO=3,36/5,6*100%=....
nHNO3=4nNO+ 10nN2O=4*0,15+10*0,1=1,6 mol
=> CM HNO3=1,6/0,5=3,2M
đề thiếu tỉ số khí nhé
nhh = 5.6/22.4 = 0.25 mol
3Mg + 8HNO3 --> 3Mg(NO3)2 + 2NO + 4H2O
x_______8x/3_________________2x/3
4Mg + 10HNO3 --> 4Mg(NO3)2 + N2O + 5H2O
y______2.5y___________________y/4
<=> x + y = 0.625
2x/3 + y/4 = 0.25
=> x = 0.225
y = 0.4
%NO = 0.225/0.25 *100% = 90%
%N2O = 10%
nHNO3 = 1.6 mol
CM HNO3 = 1.6/0.5 = 3.2M