- 2Na + 2H₂O -----> 2NaOH + H₂

- Áp dụng định luật BTKL được :

mH2O = mNaOH + mH2 - mNa = 8 + 0,2 - 4,6 = 3,6 (g)

\(n_{CO_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\\ n_{NaOH}=\dfrac{6,4}{40}=0,16\left(mol\right)\)

\(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,16}{0,7}\approx2,2\rightarrow\) Tạo muối trung hoà

PTHH: 2NaOH + CO₂ ---> Na₂CO₃ + H₂O

LTL: \(\dfrac{0,16}{2}>0,07\rightarrow\) NaOH dư

Theo pthh: \(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=2n_{CO_2}=0,07.2=0,14\left(mol\right)\\n_{Na_2CO_3}=n_{CO_2}=0,07\left(mol\right)\end{matrix}\right.\)

\(\rightarrow m_{sau.pư}=\left(0,16-0,14\right).40+0,07.106=8,22\left(g\right)\)

- Xét phần 1: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)

PTHH: 

\(2Na+2HCl\rightarrow2NaCl+H_2\)

a---------------------------->0,5a

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

b----------------------------->b

`=> 0,5a + b = 0,05 (1)`

- Xét phần 2: Đặt hệ số tỉ lệ \(\dfrac{P_2}{P_1}=k\left(k>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=ak\left(mol\right)\\n_{Mg}=bk\left(mol\right)\end{matrix}\right.\)

Ta có: \(m_{gi\text{ả}m}=m_{O\left(CuO\right)\left(p\text{ư}\right)}=8-6,72=1,28\left(g\right)\)

\(\Rightarrow n_{CuO\left(p\text{ư}\right)}=n_{O\left(p\text{ư}\right)}=\dfrac{1,28}{16}=0,08\left(mol\right)\)

PTHH: 

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

ak-------------------------->0,5ak

\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

0,08-->0,08

`=> 0,5ak = 0,08 <=> ak = 0,16 (2)`

- Xét hỗn hợp ban đầu:

Ta có: \(\dfrac{hhb\text{đ}}{P_1}=\dfrac{P_1+P_2}{P_1}=\dfrac{P_1+k.P_1}{P_1}=k+1\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=a\left(k+1\right)\left(mol\right)\\n_{Mg}=b\left(k+1\right)\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow23a\left(k+1\right)+24b\left(k+1\right)=8,2\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=0,04\\b=0,03\\k=4\end{matrix}\right.\left(TM\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,04.23}{0,04.23+0,03.24}.100\%=56,1\%\\\%m_{Mg}=100\%-56,1\%=43,9\%\end{matrix}\right.\)

a)

\(C_2H_4 + Br_2 \to C_2H_4Br_2 n_{C_2H_4} = n_{Br_2} = \dfrac{8}{160} = 0,05(mol)\\ \Rightarrow \%V_{C_2H_4} = \dfrac{0,05.22,4}{2,24} .100\%= 50\%\\ \%V_{C_2H_4} = 100\%-50\% = 50\%\)

b)

\(C_2H_4 + H_2 \xrightarrow{t^o,Ni} C_2H_6\\ n_{H_2\ pư} = n_{C_2H_6} = \dfrac{0,896}{22,4} = 0,04(mol)\\ \Rightarrow m_{H_2\ pư} = 0,04.2 = 0,08(gam)\)

\(a)\\ m_{H_2O} = m_{tăng} = 0,9\ gam\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{CuO\ pư} = n_{H_2O} = \dfrac{0,9}{18} = 0,05(mol)\\ \Rightarrow m_{CuO\ pư} = 0,05.80 = 4\ gam\\ b)\\ H = \dfrac{4}{8}.100\% = 50\%\)

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

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

Ta có: \(n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\)

Theo PT: \(n_{CuO\left(pư\right)}=n_{H_2O}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO\left(pư\right)}=0,05.80=4\left(g\right)\)

b, Ta có: \(H\%=\dfrac{4}{8}.100\%=50\%\)

Bạn tham khảo nhé!

 Phương trình : 

CuO + H₂ => Cu + H₂O

Khối lượng bình tăng lên 0,9g => \(m_{H_2O}=0,9\left(g\right)\)

=> \(n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\)

Theo pt: \(nH_2O=nCuO=\dfrac{8}{80}=0,1\left(mol\right)\)

Vậy hiệu suất phản ứng là : H = \(\dfrac{0,05}{0,1}=50\%\)

Bài 1:

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\)

\(\Rightarrow V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)

\(\Rightarrow V_{CH_4}=3,36-1,12=2,24\left(l\right)\)

Bài 2:

Ta có: \(n_{C_2H_2}+n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)

m _{dd tăng} = m_C2H2 + m_C2H4

\(\Rightarrow6,8=26n_{C_2H_2}+28n_{C_2H_4}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_2}=0,1\left(mol\right)\\n_{C_2H_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,1.22,4=2,24\left(l\right)\\V_{C_2H_4}=0,15.22,4=3,36\left(l\right)\end{matrix}\right.\)

Na2CO3 + H2SO4 -> Na2SO4 + CO2 + H2O

CO2 + 2NaOH -> Na2CO3 + H2O 

a             2a           a

CO2 + NaOH -> NaHCO3

b            b                 b

nNaOH= \(\frac{8\cdot40\%}{40}\)= 0,08 mol

Theo đề bài ta có hệ pt

2a+b=0,08

106a+84b= 5,48

=> a=0,02        b=0,04

%mNa2CO3 = \(\frac{0,02\cdot106}{5,48}\)*100% = 38,69%

%mNaHCO3 = 61,31%

nCO2 = 0,02+0,04=0,06

CMH2SO4 = 0,06/0,2=0,3 M