\(a,PTHH:CuSO_4+2KOH\rightarrow Cu\left(OH\right)_2+K_2SO_4\\ b,n_{CuSO_4}=\dfrac{80}{160}=0,5\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCO_4}=1\left(mol\right)\\ \Rightarrow V_{dd_{KOH}}=\dfrac{1}{1}=1\left(l\right)\\ c,n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,5\left(mol\right)\\ \Rightarrow m_{Cu\left(OH\right)_2}=0,5\cdot98=49\left(g\right)\)

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

a) PTHH : \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

                   0,125<-----0,25---------------------->0,125

b) \(m_{Kt}=m_{Cu\left(OH\right)2}=0,125.98=12,25\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH:

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

0,125       0,25                0,125         0,25

\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ b,Qu\text{ỳ}-t\text{í}m-ho\text{á}-\text{đ}\text{ỏ}-do-c\text{ó}-\text{ax}it-HCl\\ c,n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,1\left(mol\right)\\ m_{k\text{ết}-t\text{ủa}}=m_{BaSO_4}=233.0,1=23,3\left(g\right)\\ d,m_{\text{dd}H_2SO_4}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\\ e,m_{\text{dd}HCl}=200+200-23,3=376,7\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ C\%_{\text{dd}HCl}=\dfrac{0,2.36,5}{376,7}.100\approx1,938\%\)

Theo đề bài ta có hệ phương trình

a)

$Fe + CuSO_4 \to FeSO_4 + Cu$

Theo PTHH : $n_{Cu} = n_{CuSO_4} = 0,3.1 = 0,3(mol)$
$m_{Cu} = 0,3.64 = 19,2(gam)$

b) $n_{FeSO_4} = n_{CuSO_4} = 0,3(mol)$
$\Rightarrow m_{FeSO_4} = 0,3.152 = 45,6(gam)$

c) $FeSO_4 + 2NaOH \to Fe(OH)_2 + Na_2SO_4$
$n_{Fe(OH)_2} = n_{FeSO_4} = 0,3(mol)$
$m_{Fe(OH)_2} = 0,3.90 = 27(gam)$

a) \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

b) \(n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)

PTHH:  \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

            0,25----->0,25------->0,25---->0,5

=> \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) \(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)

d) 

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ \(\dfrac{0,5}{1}>\dfrac{0,2}{1}\) => NaOH hết, HCl dư

=> Quỳ tím chuyển màu đỏ

Số mol Br₃C₆H₂OH = 19,86: 331,0 = 0,0600 (mol)

m_phenol = 0,0600. 94,0 = 5,64 g

 = 0,2400 mol

   = 0,2400 . 46,0 = 11,05 g

Phần trăm khối lượng của etanol là 66,2% phần trăm khối lượng của phenol là 33,8%

a) 2Al + 6HCl → 2AlCl3 + 3H2

b) n Al= 8,1/27 =0,3(mol)

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4= 10,08 lít

c) n AlCl3 = n Al = 0,3 mol

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 2n H2 = 0,9(mol)

=> V dd HCl = 0,9/0,5 = 1,8(lít)

Theo đề bài ta có hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{x}{y}+\dfrac{y}{2}=\dfrac{6,72}{22,4}=0,3\\331y=16,55\end{matrix}\right.\)

Giải ra ta được :

x = 0,55

y = 0,05

=> \(mC_2H_5OH=0,55.46=25,3\left(g\right)\)

=> \(mC_6H_5OH=0,05.94=4,7\left(g\right)\)

=> \(\%m_{C_2H_5OH}=\dfrac{25,3.100}{25,3+4,7}=84,33\%\)

=> \(\%m_{C_6H_5OH}=100-84,33=15,67\%\)