k có hiệu suất hả?

\(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

Mg + 2HCl ---> MgCl₂ + H₂

0,05...0,1............0,05

\(m_{HCl}=0,1.36,5=3,65g\)

\(m_{dd\left(HCl\right)}=\dfrac{3,65.100}{14,6}=25g\)

_ trích mỗi chất 1 ít làm mẫu thử

_ cho nước vào 2 mẫu, thấy phản ứng có tỏa nhiệt là CaO: CaO + H2O ---> Ca(OH)2

_ còn lại là: Ca(OH)2

BLACKPINK vs BTS

1, Fe + 2HCl -----> FeCl₂ + H₂

2,Fe(OH)₃ + 3HCl -----> FeCl₃ + 3H₂O

3,CuO + 2HCl -----> CuCl₂ + H₂O

4,Cu+2H₂SO₄(dac,nong) --->2H₂O + SO₂+ CuSO₄

2, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{200}{1000}.2,5=0,5\left(mol\right)\)

2Al + 6HCl ----> 2AlCl₃ + 3H₂

Ta co: \(\dfrac{0,1}{2}< \dfrac{0,5}{6}\Rightarrow\) HCl dư

2Al + 6HCl ----> 2AlCl₃ + 3H₂

de: 0,1.......0,5

pu: 0,1......0,3................0,1.........0,15

spu: 0 .......0,2.................0,1........0,15

a, \(V_{H_2}=0,15.22,4=3,36l\)

b, \(m_{HCl}=1,25.200=250g\)

\(m_{dd}=250+2,7-0,15.2=252,4g\)

\(m_{HCl\left(dư\right)}=0,2.36,5=7,3g\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

\(C\%_{HCl\left(dư\right)}=\dfrac{7,3}{252,4}.100\%\approx2,89\%\)

\(C\%_{AlCl_3}=\dfrac{13,35}{252,4}.100\%\approx5,29\%\)

100ml = 0,1l

\(n_{H_2SO_4}=0,1.2,25=0,225\left(mol\right)\)

gọi x la so mol cua A

2A + xH2SO4 ----> A₂(SO₄)_x + xH2

pt: 2A(g) x (mol)

de:4,05g 0,225

Ta co: 0,45A = 4,05x

=> A = 9x

bien luan:

+ x = 1 => A = 9 => la Be mà Be k td vs

H2SO4 => loại

+ x= 2 => A = 18 (loai)

+ x= 3 => A = 27 (Lấy)

=> A là Al

\(n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\)

2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2

de: 0,15..........................0,075............0,225

b, \(V_{H2}=22,4.0,225=5,04l\)

c, \(m_{Al2\left(SO4\right)3}=0,075.342=25,65g\)

1, CTHH: Ca_xN_yO_z

\(m_{Ca}:m_N:m_O=10:7:24\)

\(\Rightarrow x:y:z=\dfrac{10}{40}:\dfrac{7}{14}:\dfrac{24}{16}\)

\(\Rightarrow x:y:z=1:2:6\)

=> CT: Ca(NO₃)₂

4, CT: C_xS_y

\(m_C:m_S=3:16\)

\(\Rightarrow x:y=\dfrac{3}{12}:\dfrac{16}{32}\)

\(\Rightarrow x:y=1:2\)

=> CT: CS₂