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

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

0,15       0,15        0,15      0,075 

a. \(m_{H_2O}=0,15.18=2,7\left(g\right)\)

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

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

0,075            0,075

Lập tỉ lệ: \(\dfrac{0,075}{2}< \dfrac{0,05}{1}\)

=> Lượng \(H_2\) sinh ra không đủ để pứ với 1,6 g \(O_2\)

\(m_{H_2O}=0,075.18=1,35\left(g\right)\)

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

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂ 

           0,15---------------->0,15---->0,075

=> \(m_{\text{dd}NaOH}=\dfrac{0,15.40}{10\%}=60\left(g\right)\)

Ta có: \(m_{\text{dd}NaOH}=m_{Na}+m_{H_2O}-m_{H_2}\)

=> \(m=m_{H_2O}=60-3,45+0,075.2=56,7\left(g\right)\)

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

PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,075}{2}< \dfrac{0,05}{1}\Rightarrow O_2\) dư, H₂ không đủ để đốt cháy hết

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

=> \(m_{s\text{ản}.ph\text{ẩm}}=m_{H_2O}=0,075.18=1,35\left(g\right)\)

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)

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

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

nNa2CO3= 25/106(mol)

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

a) nHCl=25/106 . 2= 25/53 (mol)

=> m=mddHCl={[25/53].36,5]/15%}=114,78(g)

b) nCO2= 25/106 x 22,4= 5,28(l)

c) mNaCl=25/53. 58,5=27,59(g)

mddNaCl=25+114,78- 25/106.44=129,4(g)

=>C%ddNaCl=(27,59/129,4).100=21,32%

a)

\(Zn + 2HCl \to ZnCl_2 + H_2\\ n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{19,5}{65} =0,3(mol)\\ V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{ZnCl_2} = 0,3.136 = 40,8(gam)\\ c) n_{HCl} = 2n_{Zn} = 0,6(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,6}{2} = 0,3(lít)\\ d) 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{1}{2} V_{H_2} = 3,36(lít)\\ V_{không\ khí} = \dfrac{V_{O_2}}{20\%}= \dfrac{3,36}{20\%} = 16,8(lít)\)

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

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

\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)