Решите электронно-ионное уравнение zn+hn03-> zn(bo3)2+no2+h2o

cu+> cu(no3)2+2no2+2h2o

v=v_{m} * n; n=m/m; v=v_{m} * m;

a) v=22.4 * 0.6 = 13.44л;

б) n=5; v=22.4 * 5 = 112л;

в) v=22.4 *4.5 = 100.8л

dano:

m(al)=3g

v(

4al+3o2=2al2o3

n(al)=3g/27g/mol

n(al)=0.1mol

n(o2)=0.1*3/4=0.075mol

v(o2)=0.075mol*22.4l/mol=1.68l

      zn+4hn03-> zn(no3)2+2no2+2h2o