Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=C_M\cdot V=1\cdot0,05=0,05\left(mol\right)\)
Nhận xét: \(1< \dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,06}{0,05}< 2\rightarrow\) Phương trình tạo ra 2 muối
Đặt \(\left\{{}\begin{matrix}n_{CaCO_3}=x\left(mol\right)\\n_{NaHCO_3}=y\left(mol\right)\end{matrix}\right.\) và x; y > 0
a) PHTT
Ca(OH)2 + CO2 \(\rightarrow\) CaCO3
CaCO3 + CO2 + H2O \(\rightarrow\) Ca(HCO3)2
b)
Ca(OH)2 + CO2 \(\rightarrow\) CaCO3
x \(\leftarrow\) x \(\leftarrow\) x (mol)
Ca(OH)2 + 2CO2 \(\rightarrow\) Ca(HCO3)2
y \(\leftarrow\) 2y \(\leftarrow\) y (mol)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,05\\x+2y=0,06\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,04\left(mol\right)\\y=0,01\left(mol\right)\end{matrix}\right.\)
Nhận xét, \(n_{CaCO_3\left(ht\right)}=n_{Ca\left(HCO_3\right)_2}=0,01\left(mol\right)\)
\(\Rightarrow m_{CaCO_3\left(ht\right)}=n\cdot M=0,01\cdot100=1\left(g\right)\)
c)
Để kết tủa cực đại thì
\(\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=1\Leftrightarrow n_{CO_2}=n_{Ca\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{CO_2}=n\cdot22,4=0,05\cdot22,4=1,12\left(l\right)\)
\(a)n_{CO2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
\(n_{Ca\left(OH\right)2}=0,05.1=0,05\left(mol\right)\)
\(\dfrac{n_{CO2}}{n_{Ca\left(OH\right)2}}=1,2\Rightarrow\)Tạo 2 muối
\(CO2+Ca\left(OH\right)2\rightarrow CaCO3+H2O\)
0,05-------<0,05(mol)
\(CO2+CaCO3+H2O\rightarrow Ca\left(HCO3\right)2\)
0,01-------->0,01(mol)
\(b)m_{CaCO3}=0,01.100=1\left(g\right)\)
c) Để đạt kết tủa cực tại thì\(\dfrac{n_{Ca\left(OH\right)2}}{n_{CO2}}=2\)
Mà \(n_{Ca\left(OH\right)2}=0,05\left(mol\right)\)
\(\Rightarrow n_{CO2}=0,025\left(mol\right)\)
\(\Rightarrow V_{CO2}=0,025.22,4=0,56\left(l\right)\)
\(a.n_{CO_2}=\dfrac{0,672}{22,4}=0,03mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
\(n_{CO_2}=n_{Ca\left(OH\right)_2}=n_{CaCO_3}=0,03mol\\ m_{CaCO_3}=0,03.100=3g\\ b.V_{ddCa\left(OH\right)_2}=\dfrac{0,03}{1,5}=0,02l\)
\(n_{Ca\left(OH\right)_2}=0,2\cdot1=0,2mol\)
\(Ca\left(ỌH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
0,2 0,2 0,2
\(m_{CaCO_3}=0,2\cdot\left(40+12+3\cdot16\right)=20\left(g\right)\)
\(V_{CO_2}=0,2\cdot22,4=4,48\left(l\right)\)
Bài 1 :
$n_{CO_2} = \dfrac{3,136}{22,4} = 0,14(mol)$
$n_{Ca(OH)_2} = 0,8.0,1 = 0,08(mol)$
CO2 + Ca(OH)2 → CaCO3 + H2O
0,08.......0,08...........0,08........................(mol)
CaCO3 + CO2 + H2O → Ca(HCO3)2
0,06........0,06........................................(mol)
Suy ra : $m_{CaCO_3} = (0,08 - 0,06).100 = 2(gam)$
Bài 2 :
$n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol) ; n_{NaOH} = 0,1.1,5 = 0,15(mol)$
2NaOH + CO2 → Na2CO3 + H2O
0,15........0,075.......0,075....................(mol)
Na2CO3 + CO2 + H2O → 2NaHCO3
0,025........0,025...................0,05..............(mol)
Suy ra:
$C_{M_{NaHCO_3}} = \dfrac{0,05}{0,1} = 0,5M$
$C_{M_{Na_2CO_3}} = \dfrac{0,075 - 0,025}{0,1} = 0,5M$
b)
$NaOH + HCl \to NaCl + H_2O$
$n_{HCl} = n_{NaOH} = 0,15(mol)$
$m_{dd\ HCl} = \dfrac{0,15.36,5}{25\%} = 21,9(gam)$
1.
\(n_{CO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0.075\left(mol\right)\)
\(T=\dfrac{0.1}{0.075}=1.33\)
=> Tạo ra 2 muối
\(n_{CaCO_3}=a\left(mol\right),n_{Ca\left(HCO_3\right)_2}=b\left(mol\right)\)
Khi đó :
\(a+b=0.075\)
\(a+2b=0.1\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.05\\b=0.025\end{matrix}\right.\)
\(m_{sp}=0.05\cdot100+0.025\cdot162=9.05\left(g\right)\)
2.
\(n_{CO_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0.2\cdot0.2=0.04\left(mol\right)\)
\(T=\dfrac{0.005}{0.04}=1.25\)
=> Tạo ra 2 muối
\(n_{BaCO_3}=a\left(mol\right),n_{Ba\left(HCO_3\right)_2}=b\left(mol\right)\)
Ta có :
\(a+b=0.04\)
\(a+2b=0.05\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.03\\b=0.01\end{matrix}\right.\)
\(m_{BaCO_3}=0.03\cdot197=5.91\left(g\right)\)
Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
______0,2_____0,2_________0,2 (mol)
b, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4M\)
c, \(m_{CaCO_3}=0,2.100=20\left(g\right)\)
Bạn tham khảo nhé!
\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ a,n_{O_2}=3.0,5=1,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\\ V_{kk\left(đktc\right)}=33,6.5=168\left(l\right)\\ b,n_{CO_2}=n_{H_2O}=2.0,5=1\left(mol\right)\\ m_{CO_2}=44.1=44\left(g\right);m_{H_2O}=18.1=18\left(g\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{Ca\left(OH\right)_2}=n_{CO_2}=1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=1.74=74\left(g\right)\\ m_{ddCa\left(OH\right)_2}=\dfrac{74.100}{10}=740\left(g\right)\)
nCO2=\(\dfrac{6,72}{22,4}\)=0,3 m0l
mct Ca(OH)2=\(\dfrac{148.10}{100}\)=14,8gam ; nCa(OH)2=\(\dfrac{14,8}{74}\)=0,2 mol
a............................ CO2 + Ca(OH)2 ----> CaCO3 + H2O
số mol trước pư......0,3........0,2.................................................mol
số mol pư...............0,2.........0,2..................................................mol
số mol sau pư.........0,1...........0.......................0,2...........0,2.......mol
c. Để có kl kết tủa cực đại thì nCa[OH]2= 0,3 mol
=> mCa[OH]2=0,3.74=22,2 gam
Bài này em làm chưa chính xác rồi. Sản phẩm tạo thành chứa 2 muối
\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{O_2}=3.0,4=1,2\left(mol\right);n_{CO_2}=0,4.2=0,8\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=22,4.1,2=26,88\left(l\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.26,88=134,4\left(l\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,8\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.0,8=80\left(g\right)\)
\(n_{CO_2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,05\times1=0,05\left(mol\right)\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\) (1)
\(CaCO_3+H_2O+CO_2\rightarrow Ca\left(HCO_3\right)_2\) (2)
Theo (1) thấy \(Ca\left(OH\right)_2\) hết \(\Rightarrow m_{Ca\left(OH\right)_2}=0,05\times74=3,7\left(g\right)\)
\(n_{CO_2pu}=n_{Ca\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,05\times22,4=1,12\left(mol\right)\)
Bạn ghi đề đúng rồi mình giải cho nhé.