\(NaOOC-COONa\rightarrow\left(COO^-\right)_2+2Na^+\)

FDDDDDDDDDDDDDDDDDDDDDDDDDDDĐ 

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cứu mẹ

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\(n_{H_2}=\frac{3,92}{22,4}=0,175mol\)

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

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

b. Đặt \(\hept{\begin{cases}x\left(mol\right)=n_{Mg}\\y\left(mol\right)=n_{Al}\end{cases}}\)

\(\rightarrow24x+27y=3,75\left(1\right)\)

Theo phương trình \(n_{Mg}+1,5n_{Al}=n_{H_2}=0,175\)

\(\rightarrow x+1,5y=0,175\left(2\right)\)

Từ (1) và (2) \(\rightarrow\hept{\begin{cases}x=0,1mol\\y=0,05mol\end{cases}}\)

\(\rightarrow m_{Mg}=0,1.24=2,4g\)

\(\rightarrow m_{Al}=3,75-2,4=1,35g\)

FDDDDDDDDDDDDDDDDDDDDDDDDDDDĐ 

1.

a. \(Fe_2O_3+3H_2\rightarrow^{t^o}2Fe+3H_2O\)

b. \(HgO+H_2\rightarrow^{t^o}Hg+H_2O\)

c. \(PbO+H_2\rightarrow^{t^o}Pb+H_2O\)

d. \(Ag_2O+H_2\rightarrow^{t^o}2Ag+H_2O\)

2.

\(4H_2+Fe_3O_4\rightarrow^{t^o}3Fe+4H_2O\)

\(n_{Fe}=\frac{16,8}{56}=0,3mol\)

a. Theo phương trình \(n_{Fe_3O_4}=n_{Fe}.\frac{1}{3}=0,3.\frac{1}{3}=0,1mol\)

\(\rightarrow m_{Fe_3O_4}=0,1.\left(56.3+16.4\right)=23,2g\)

b. Theo phương trình \(n_{H_2}=n_{Fe}.\frac{3}{4}=0,3.\frac{4}{3}=0,4mol\)

\(\rightarrow V_{H_2}=0,4.22,4=8,96l\)

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Nguồn ;mạng

Gửi bài r

~HT~

\(200ml=0,2l\)

\(n_{Ba\left(OH\right)_2}=C_M.V=0,45.0,2=0,09mol\)

PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

                  0,09                   0,09                                          mol

\(\rightarrow nH_2SO_4=n_{Ba\left(OH\right)_2}=0,09mol\)

\(\rightarrow V_{H_2SO_4}=\frac{n}{C_M}=\frac{0,09}{0,3}=0,3l\)

Vậy chọn A.

Khối lượng dung dịch NaOH 8% cần dùng để tác dụng hết với 200g dung dịch H2SO4 19,6% *   a. 39,2g   b. 32g   c. 400g   d. 600g