Bài 1: Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____1_____2_____________1 (mol)

a, Ta có: \(V_{H_2}=1.22,4=22,4\left(l\right)\)

b, Ta có: \(m_{HCl}=2.36,5=73\left(g\right)\)

Bài 2: Ta có: \(n_{CuO}=\dfrac{100}{80}=1,25\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

___1,25_______1,25__1,25 (mol)

a, Ta có: \(m_{Cu}=1,25.64=80\left(g\right)\)

b, \(m_{H_2O}=1,25.18=22,5\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Na}=\dfrac{9,2}{23}=0,4mol\)

\(Na+CH_3COOH\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)

0,4             0,4                                        ( mol )

\(m_{CH_3COOH}=0,4.60=24g\)

2CH3COOH+Na->2CH3COONa+H2

0,8---------------0,4

n Na=0,4 mol

=>m CH3COOH=0,8.60=48g

$a\big)$

$n_{Na}=\dfrac{4,6}{23}=0,2(mol)$

$CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2$

Theo PT: $n_{CH_3COOH}=n_{Na}=0,2(mol)$

$\to m_{CH_3COOH}=0,2.60=12(g)$

$b\big)$

Theo PT: $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

giúp tuiii

Ptpư    Zn + 2HCl  ---> ZnCl2 + H2

nH2 = 1 mol

mH2 = 2(g)

Áp dụg ĐLBTKL 

mHCl = mZnCl2 +mH2 - mZn = 136 +2-65= 73(g)

ko hjeu đề bài.........bn sửa lại đj,mk lm cho

a, \(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\)

\(\Rightarrow a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\)

b, \(m_{dd}=\dfrac{15}{2\%}=750\left(g\right)\)

\(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\\ NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\\a, n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\\ a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\\ b,m_{ddCH_3COOH}=\dfrac{15.100}{2}=750\left(g\right)\)

a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

b, Phần này đề hỏi tính khối lượng gì bạn nhỉ?

c, \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{Zn}=1\left(mol\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{1.60}{36\%}=\dfrac{500}{3}\left(g\right)\)

a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

_____0,1--->0,2------->0,1---->0,1

=> m_HCl = 0,2.36,5 = 7,3(g)

b) m_MgCl2 = 0,1.95 = 9,5 (g)

c) V_H2 = 0,1.22,4 = 2,24(l)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

0,1         0,2

a. \(V_{CH_3COOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)

b. \(CH_3COOH+C_2H_5OH⇌\left(H_2SO_{4đ},t^o\right)CH_3COOC_2H_5+H_2O\)

          0,2                                                             0,2

Với H% = 80

\(m_{CH_3COOC_2H_5}=\dfrac{0,2.88.80}{100}=14,08\left(g\right)\)