Ống 1 : Không phản ứng

Ống 2 : CaO + H₂ -> Ca + H₂O

            0,01               0,01

Ống 3 : PbO + H₂ -> Pb + H₂O

             0,02            0,02

Ống 4 : Fe₂O₃ + 3H₂ -> 2Fe + 3H₂O

              0,01                   0,02

Ống 5 : Na₂O + H₂ -> 2NaOH

             0,06                 0,12

Khối lượng các chất rắn thu được là:

\(m_{Ca}=0,01.40=0,4\left(g\right)\)

\(m_{Pb}=0,02.207=4,14\left(g\right)\)

\(m_{Fe}=0,02.56=1,12\left(g\right)\)

\(m_{NaOH}=0,12.40=4,8\left(g\right)\)

a)\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)

\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)

\(\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)

\(\frac{1}{3}.x=\frac{35}{12}\)

\(x=\frac{35}{12}:\frac{1}{3}\)

\(x=\frac{35}{12}.\frac{3}{1}\)

\(x=\frac{35}{4}\)

b)\(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)

\(4,5.\left(0,1.x\right)=2,25.0,3\)

\(4,5.\left(0,1.x\right)=0,675\)

\(0,1.x=0,675:4,5\)

\(0,1.x=0,15\)

\(x=0,15:0,1\)

\(x=1.5\)

c)\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)

\(\left(\frac{1}{4}.x\right).2=8.0,02\)

\(\left(\frac{1}{4}.x\right).2=0,16\)

\(\frac{1}{4}.x=0,16:2\)

\(\frac{1}{4}.x=0,08\)

\(x=0,08:\frac{1}{4}\)

\(x=0,32\)

d)\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)

\(3.\left(6.x\right)=\frac{3}{4}.\frac{9}{4}\)

\(3.\left(6.x\right)=\frac{27}{16}\)

\(6.x=\frac{27}{16}:3\)

\(6.x=\frac{9}{16}\)

\(x=\frac{9}{16}:6\)

\(x=\frac{3}{32}\)

HỌC TỐT ^^

= 65,384 bạn nhé !

Nhớ k cho mình nha ! Thank you very much !

a. 

Bảo toàn điện tích : 

\(n_{NO_3^-}=0.01\cdot3+0.02-0.02\cdot2=0.01\left(mol\right)\)

b.

\(n_{Ba\left(OH\right)_2}=0.1\cdot0.3=0.03\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

\(0.02.....0.02..........0.02\)

\(Ba^{2+}dư\)

\(m_{BaSO_4}=0.02\cdot233=4.66\left(g\right)\)

\(NH_4^++OH^-\rightarrow NH_3+H_2O\)

\(0.02.......0.02........0.02\)

\(V_{NH_3}=0.02\cdot22.4=0.448\left(l\right)\)

\(OH^-dư\)