a)

\(n_{Na} = \dfrac{3,45}{23} = 0,15(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\\)

Theo PTHH : 

\(n_{NaOH} = n_{Na} = 0,15(mol)\\ m_{dung\ dịch\ NaOH} = \dfrac{0,15.40}{10\%} = 60(gam)\\ m_{H_2O\ đã\ dùng} = m = m_{dd\ NaOH} - m_{NaOH} = 60 -0,15.40 = 54(gam)\)

b)

\(n_{H_2} = \dfrac{n_{Na}}{2} = 0,075(mol)\\ n_{O_2} =\dfrac{1,6}{32} = 0,05(mol)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\)

Ta thấy : \(\dfrac{n_{H_2}}{2} = 0,0375 < n_{O_2} \Rightarrow\) O₂ dư. Do đó, lượng H₂ sinh ra không đủ phản ứng hết với 1,6 gam oxi.

\(n_{H_2O} = n_{H_2} = 0,075(mol)\\ \Rightarrow m_{H_2O} = 0,075.18 = 1,35(gam)\)

a) PTHH: 2Na + 2H₂O \(\rightarrow\) 2NaOH + H₂\(\uparrow\)(1)
n_Na = \(\frac{3,45}{23}=0,15\left(mol\right)\)
Theo PT(1): n_{\(H_2O\)} = n_Na = 0,15 (mol)
=> m_{\(H_2O\)} = 0,15.18 = 2,7 (g) = m
b) PTHH: 2H₂ + O₂ \(\underrightarrow{t^o}\) 2H₂O(2)
n_{\(O_2\)} = \(\frac{1,6}{32}=0,05\left(mol\right)\)
Theo PT(1): n_{\(H_2\)} = \(\frac{1}{2}n_{Na}\) = \(\frac{1}{2}.0,15=0,075\left(mol\right)\)= n_{\(H_2\)(2)}
Ta có tỉ lệ: \(\frac{n_{H_2}}{2}=\frac{0,075}{2}=0,0375< \frac{n_{O_2}}{1}=0,05\)
=> H₂ hết, O₂ dư => lượng hidro sinh ra đủ để pứ với 1,6g oxi
=> Tính số mol các chất cần tìm theo H₂
Theo PT(2): n_{\(H_2O\)} = n_{\(H_2\)} = 0,075(mol)
=> m_{\(H_2O\)} = 0,075.18 = 1,35(g)

a) PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)

\(n_{Na}=\frac{3,45}{23}=0,15\left(mol\right)\)

Theo pt \(\Rightarrow n_{NaOH}=0,15\cdot40=6\left(mol\right)\)

\(\Rightarrow\) m_dd NaOH = \(\frac{6\cdot100\%}{10\%}=60\left(g\right)\)

Áp dụng ĐLBTKL:

\(\frac{6}{m+3,45}\cdot100\%=10\%\Leftrightarrow6=0,1m+3,45\)

\(\Rightarrow m=56,7\left(g\right)\)

b) PTHH: \(2H_2+O_2\rightarrow2H_2O\)

\(n_{H_2}=\frac{0,15}{2}=0,075\left(mol\right)\)

\(n_{O_2}=\frac{1,6}{32}=0,05\left(mol\right)\)

Lập tỉ lệ : \(\frac{0,075}{2}< 0,05\)

\(\Rightarrow\) O₂ dư, kết luận lượng H₂ sinh ra không đủ để phản ứng với 1,6g O₂.

Theo pt \(\Rightarrow n_{H_2O}=n_{H_2}=0,075\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,075\cdot18=1,35\left(g\right)\)

a) 2Na+2H2O--->2NaOH+H2

n Na=3,45/23=0,15(mol)

n NaOH=n Na=0,15(mol)

m NaOH=0,15.40=6(g)

m dd NaOH=6.100/10=60(g)

m nước=m dd-m NaOH=60-6=54(g)

b) 2H2+O2--->2H2O

n H2=1/2n Na=0,075(mol)

n O2=1,6/32=00,05(mol)

--->O2 dư...Lượng H2 k đủ phản ứng với O2

n H2O=n H2=0,075(mol)

m H2O=0,075.18=1,35(g)

Bài Đầu tiên

a, PTHH : 2Na + 2H₂O -> 2NaOH + H₂

n_Na = \(\dfrac{3,45}{23}=0,15\left(mol\right)\)

Ta có , m_{dung dịch} = m + 3,45 (g)

m_{dung dịch bazo} = m_NaOH ( cái này khác với dung dịch sau phản ứng nhé )

Theo PTHH , n_Na= n_NaOH= 0,15 (mol)

=> m_{NaOH =} 0,15 . 40 =6(g)

Theo bài ra ta có hệ phương trình sau :

\(\dfrac{6}{m+3,45}.100\%=10\%\Leftrightarrow6=0,1m+0,345\)

=> m = 56,55 (g)

b, n_H2= 0,15 /2 = 0,075 (mol)

PTHH : 2H₂ + O₂ -> 2H₂O

n_O2=1,6/32 =0,05 (mol)

Vì 0,075/2 < 0,05 => Oxi dư => H₂ thiếu

=> lượng Hidro không đủ

Bài 5:

m_HCl = \(\dfrac{3,65\times100}{100}=3,65\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{3,65}{36,5}=0,1\)mol

Pt: 2Al + 6HCl --> 2AlCl₃ + 3H₂

.....Fe + 2HCl --> FeCl₂ + H₂

.....Mg + 2HCl --> MgCl₂ + H₂

Theo pt ta có: n_H2 = \(\dfrac{1}{2}\)n_HCl = \(\dfrac{1}{2}.0,1=0,05\) mol

=> V_H2 = 0,05 . 22,4 = 1,12 (lít)

m_{muối khan} = m_{kim loại} + m_{gốc axit} = 1,34 + 0,1 . 35,5 = 4,89 (g)

Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

a. PTHH: 2Na + 2H₂O ---> 2NaOH + H₂↑

b. Ta có: \(n_{H_2O}=\dfrac{97,8}{18}=5,43\left(mol\right)\)

Ta thấy: \(\dfrac{0,1}{2}< \dfrac{5,43}{2}\)

=> H₂O dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(V_{H_2}=0,05.22,4=1,12\left(lít\right)\)

c. Ta có: \(m_{dd_{NaOH}}=2,3+97,8=100,1\left(g\right)\)

Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)

=> \(m_{NaOH}=0,1.40=4\left(g\right)\)

=> \(C_{\%_{NaOH}}=\dfrac{4}{100,1}.100\%=3,996\%\)

a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(m_{HCl}=200.14,6\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,8}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

c, \(n_{HCl\left(pư\right)}=2n_{Zn}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 19,5 + 200 - 0,3.2 = 218,9 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{218,9}.100\%\approx3,33\%\\C\%_{ZnCl_2}=\dfrac{40,8}{218,9}.100\%\approx18,64\%\end{matrix}\right.\)

\(a)n_{Zn}=\dfrac{19,5}{65}=0,3mol\\ n_{HCl}=\dfrac{200.14,6}{100.36,5}=0,8mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,3mol\\ V_{H_2}=0,3.22,4=6,72l\\ b)m_{ZnCl_2}=0,3.136=40,8g\\ c)n_{HCl.pư}=0,3.2=0,6mol\\ C_{\%ZnCl_2}=\dfrac{40,8}{200+19,5-0,3.2}\cdot100=18,64\%\\ C_{\%HCl.dư}=\dfrac{\left(0,8-0,6\right).36,5}{200+19,5-0,3.2}\cdot100=3,33\%\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a. PTHH: \(Zn+H_2SO_4--->ZnSO_4+H_2\uparrow\left(1\right)\)

b. Theo PT₍₁₎: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=65.0,3=19,5\left(g\right)\)

c. Theo PT₍₁₎: \(n_{H_2SO_4}=n_{Zn}=0,3\left(mol\right)\)

Đổi 300ml = 0,3 lít

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)

d. PTHH: \(2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\left(2\right)\)

Theo PT₍₂₎: \(n_{NaOH}=2.n_{H_2SO_4}=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)

\(\Rightarrow m_{dd_{NaOH}}=\dfrac{24.100\%}{20\%}=120\left(g\right)\)

a, \(n_{Na}=\dfrac{3,45}{23}=0,15\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Theo PT: \(n_{NaOH}=n_{Na}=0,15\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

b, \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,075\left(mol\right)\)

\(n_{O_2}=\dfrac{0,96}{32}=0,03\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,075}{2}>\dfrac{0,03}{1}\), ta được H₂ dư.

Theo PT: \(n_{H_2O}=2n_{O_2}=0,06\left(mol\right)\Rightarrow m_{H_2O}=0,06.18=1,08\left(g\right)\)

a) PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{20\%}=147\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{147}{1,14}\approx128,95\left(ml\right)\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)=n_{FeSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\V_{H_2}=0,3\cdot24,76=7,428\left(l\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=163,2\left(g\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{163,2}\cdot100\%\approx27,94\%\)