\(a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{H_2}=\dfrac{16,8}{22,4}=0,74(mol)\\ \Rightarrow n_{Fe}=0,75(mol)\\ \Rightarrow m_{Fe}=0,75.56=42(g)\\ c,n_{H_2SO_4}=\dfrac{245.10\%}{100\%.98}=0,25(mol)\)

Vì \(\dfrac{n_{Fe}}{1}>\dfrac{n_{H_2SO_4}}{1}\) nên \(Fe\) dư

\(n_{Fe(dư)}=0,75-0,25=0,5(mol)\\ \Rightarrow m_{Fe(dư)}=0,5.56=28(g)\)

a)    Fe + H₂SO₄ → FeSO₄ + H₂

_{b)  Ta có :   nH2 = \(\dfrac{16,8}{22,4}\)} = 0,75 (mol)

⇒ n_Fe= 0,75.56 = 42(gam)

Ta có: \(n_{H_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

a. \(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\uparrow\)

b. Theo PT: \(n_{Fe}=n_{H_2}=0,75\left(mol\right)\)

\(\Rightarrow m_{Fe}=56.0,75=42\left(g\right)\)

\(a.PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Từ PTHH trên ta có:

1 mol Mg tác dụng với 2 mol HCl sinh ra 1 mol MgCl₂ và 1 mol H₂ 

=> 0,1 mol Mg tác dụng với 0,2 mol HCl sinh ra 0,1 mol MgCl₂ và 0,1 mol H₂ 

\(\Rightarrow m_{HCl}=36,5.0,2=7,3\left(g\right)\)

\(c.\Rightarrow m_{H_2}=0,1.2=0,2\left(g\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(d.C_1:\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=m_{Mg}+m_{HCl}-m_{H_2}=2,4+7,3-0,2=9,5\left(g\right)\)

\(C_2:Từ.PTHH.trên.ta.có:\\ n_{MgCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.4.......0.8....................0.4\)

\(m_{HCl}=0.8\cdot36.5=29.2\left(g\right)\)

\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)

a) $Zn + H_2SO_4 \to ZnSO_4 + H_2$

b) Theo PTHH :

$n_{H_2} = n_{ZnSO_4} = n_{Zn} = \dfrac{32,5}{65} = 0,5(mol)$

$V_{H_2} = 0,5.22,4 = 11,2(lít)$

$m_{ZnSO_4} = 0,5.161 = 80,5(gam)$

c) 

$n_{H_2SO_4} = n_{Zn} = 0,5(mol)$

$C\%_{H_2SO_4} = \dfrac{0,5.98}{168,5}.100\% = 29,08\%$

Số mol: nCO2 =  = 0,07 mol; 

Phương trình hóa học:

                        2NaOH + CO2 → Na2CO3 + H2O

                                         0,07------0,07

b)Khối lượng muối Na2CO3 tạo thành là: 0,07×106 = 7,42 g.

Na2CO3+2HCl->2NaCl+H2O+CO2

0,07--------0,14

=>m HCl=0,14.36,5=5,11g

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a. PTHH:

\(Mg+2HCl--->MgCl_2+H_2\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

b. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)

\(\Rightarrow m_{MgO}=8-4,8=3,2\left(g\right)\)

c. Ta có: \(n_{hh}=0,2+\dfrac{3,2}{40}=0,28\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,28=0,56\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,56.36,5=20,44\left(g\right)\)

ai làm giúp mình với

a)

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

b) $n_{CH_4} = \dfrac{28}{22,4} = 1,25(mol)$

$n_{CO_2} = n_{CH_4} = 1,25(mol)$
$m_{CO_2} = 1,25.44 = 55(gam)$

c) $V_{O_2} =2 V_{CH_4} = 56(lít)$

\(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8(mol)\\ PTHH:2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2\uparrow+H_2\uparrow\\ \Rightarrow n_{NaOH}=n_{NaCl}=2n_{Cl_2}=1,6(mol)\\ \Rightarrow m_{NaOH}=1,6.40=64(g)\\ n_{NaCl}=1,6.58,5=93,6(g)\)