Câu 1

 \(m_{HNO_3}=0,3.63=18,9\left(g\right)\)

\(m_{CuSO_4}=1,5.160=240\left(g\right)\)

\(m_{AlCl_3}=2.133,5=267\left(g\right)\)

Câu 2

a) \(V_{N_2}=3.22,4=67,2\left(l\right)\)

\(V_{H_2}=0,45.22,4=10,08\left(l\right)\)

\(V_{O_2}=0,55.22,4=12,32\left(l\right)\)

b) \(V_{hh}=\left(0,25+0,75\right).22,4=22,4\left(l\right)\)

\(a.n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right);n_{O_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\\ V_{hh}=\left(0,5+1,5+0,1+0,1\right).22,4=49,28\left(l\right)\\ b.m_{hh}=0,5.28+1,5.2+4,4+0,1.32=24,6\left(g\right)\)

a, V_{N\(_2\)} ( đktc ) = 0,5 . 22,4 = 11,2 lít

V_{H\(_2\)}  = 1,5 . 22,4 = 33,6 lít

\(n_{CO_2}=\dfrac{4,4}{44}=0,1\) ( mol )

=> \(V_{CO_2}=0,1.22,4=2,24\) ( lít )

\(n_{O_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\) ( mol )

=> V_{\(O_2\)} = 0,1 .22,4 = 2,24 lít

=> V_hh = 11,2 + 33,6 + 2,24 + 2,24 = 49,28 lít

b, \(m_{N_2}=0,5.28=14\) ( g )

\(m_{H_2}=1,5.2=3\) ( g )

\(m_{CO_2}=0,1.44=4,4\) ( g )

\(m_{O_2}=0,1.32=3,2\) (g)

\(m_{hh}=14+3+4,4+3,2=24,6\) ( g )

m H 2 = n H 2 . M H 2  = 2.2 = 4(g) → V H 2 = n H 2 .22,4 = 2.22,4 = 44,8(l)

m O 2 = n O 2 . M O 2  = 1,5.32 = 48(g)  → V O 2 = n O 2 .22,4 = 1,5.22,4 = 33,6(l)

m C O 2 = n C O 2 . M C O 2  = 1,15.44 = 50,6(g)  → V C O 2 = n C O 2 .22,4 = 1,15.22,4 = 25,76(l)

m C H 4 = n C H 4 . M C H 4 = 1,15.16 = 18,4(g)  → V C H 4 = n C H 4 .22,4 = 1,15.22,4 = 25,76(l)

mH2 = 0,5 . 2 = 1 (g)

mO2 = 0,8 . 32 = 25,6 (g)

mCO2 = 2 . 44 = 88 (g)

1. \(m_{H_2}=0,5.2=1\left(g\right)\)

2.\(m_{O_2}=0,8.32=25,6\left(g\right)\)

3. \(m_{CO_2}=2.44=88\left(g\right)\)