Đáp án:

\(5/\\ a,\ m_{BaSO_4}=23,3\ g.\\ b,\ C\%_{HCl}=1,49\%\\ C\%_{H_2SO_4}(dư)=2,6\%\\ 9/\\ R:\ Fe\)

Giải thích các bước giải:

\(5/\\ a,\ PTHH:H_2SO_4+BaCl_2\to BaSO_4↓+2HCl\\ n_{H_2SO_4}=\dfrac{114\times 20\%}{98}≈0,23\ mol.\\ n_{BaCl_2}=\dfrac{400\times 5,2\%}{208}=0,1\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,23}{1}>\dfrac{0,1}{1}\\ ⇒H_2SO_4\ dư.\\ Theo\ pt:\ n_{BaSO_4}=n_{BaCl_2}=0,1\ mol.\\ ⇒m_{BaSO_4}=0,1\times 233=23,3\ g.\\ b,\ m_{\text{dd spư}}=114+400-23,3=490,7\ g.\\ n_{H_2SO_4}(dư)=0,23-0,1=0,13\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{BaCl_2}=0,2\ mol.\\ ⇒C\%_{HCl}=\dfrac{0,2\times 36,5}{490,7}\times 100\%=1,49\%\\ C\%_{H_2SO_4}(dư)=\dfrac{0,13\times 98}{490,7}\times 100\%=2,6\%\\ 9/\\ PTHH:R_2O_3+3H_2SO_4\to R_2(SO_4)_3+3H_2O\\ n_{H_2SO_4}=0,25\times 0,3=0,075\ mol.\\ Theo\ pt:\ n_{R_2O_3}=\dfrac{1}{3}n_{H_2SO_4}=0,025\ mol.\\ ⇒M_{R_2O_3}=\dfrac{4}{0,025}=160\ g/mol.\\ ⇒2M_{R}+3.16=160\\ ⇔2M_{R}=160-48\\ ⇔M_{R}=56\ g/mol.\\ \Rightarrow \text{R là Sắt (Fe)}\)

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Đáp án:

Giải thích các bước giải:

 Bài 5:

`H_2SO_4+BaCl_2→BaSO_4↓+2HCl`

a) `n_{H_2SO_4}=\frac{114.20%}{98} \approx 0,2\ mol`

`n_{BaCl_2}=\frac{400.5,2%}{208}=0,1\ mol`

Xét tỉ lệ : `\frac{n_{H_2SO_4}}{1}>\frac{n_{BaCl_2}}{1}`

`⇒ H_2SO_4` dư, `BaCl_2` hết 

`n_{BaSO_4}=n_{BaCl_2}=0,1\ mol`

`m_{BaSO_4}=0,1.233=23,3\ gam`

b) `m_{dd\ X}=m_{dd\ H_2SO_4}+m_{dd\ BaCl_2}-m_{BaSO_4}`

`m_{dd\ X}=114+400-23,3`

`m_{dd\ X}=490,7\ gam`

`C%_{HCl}=\frac{0,2.36,5}{490,7}.100% \approx 1,49%`

`C%_{H_2SO_4\ dư}=\frac{0,1.98}{490,7}.100 \approx 2%`

Bài 9:

`R_2O_3+3H_2SO_4→R_2(SO_4)_3+3H_2O`

`n_{H_2SO_4}=0,075\ mol`

Theo `PT` ta có:

`n_{R_2O_3}=\frac{1}{3}.n_{H_2SO_4}`

`⇔ \frac{4}{2R+48}=0,025`

`⇔ 0,025.(2R+48)=4`

`⇔ 0,05R+1,2=4`

`⇔ 0,05R=2,8`

`⇔ R=56` (g/mol) (nhận)

`⇒ R` là `Fe` (Sắt)

Vậy CTHH là `Fe_2O_3`