Cho 2,4 gam Mg vào 109,5 gam dung dịch acid hydrochloric 10% đến khi phản ứng xảy ra hoàn toàn, thu được dung dịch A và có V lít khí thoát ra (dkc). a) Viết PTPƯ, tính V b) Dung dịch A chứa chất tan nào? Tính nồng độ C% của dung dịch A.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ m_{HCl}=\dfrac{109,5\cdot10\%}{100\%}=10,95\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \text{Vì }\dfrac{n_{Mg}}{1}< \dfrac{n_{HCl}}{2}\text{ nên sau p/ứ }HCl\text{ dư}\\ \Rightarrow n_{H_2}=0,1\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\)
\(b,n_{MgCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{MgCl_2}}=0,1\cdot95=9,5\left(g\right)\\ m_{H_2}=0,1\cdot2=0,2\left(mol\right)\\ m_{dd_{MgCl_2}}=2,4+109,5-0,2=111,7\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\)
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)
\(n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)
PTHH: 2K + 2H2O --> 2KOH + H2
_____0,15------------->0,15-->0,075
=> VH2 = 0,075.22,4 =1,68(l)
mdd = 5,85 + 100 - 0,075.2 = 105,7(g)
=> \(C\%=\dfrac{0,15.56}{105,7}.100\%=7,95\%\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)
\(m_{CuCl_2}=\dfrac{270\cdot10\%}{100\%}=27g\Rightarrow n_{CuCl_2}=0,2mol\)
\(Fe+CuCl_2\rightarrow FeCl_2+Cu\)
0,15 0,2 0,15 0,15
\(a=m_{Cu}=0,15\cdot64=9,6g\)
\(m_{FeCl_2}=0,15\cdot127=19,05g\)
\(m_{ddFeCl_2}=8,4+270-0,15\cdot64=268,8g\)
\(C\%=\dfrac{19,05}{268,8}\cdot100\%=7,09\%\)
Đặt hóa trị của M là x(x>0)
\(n_{O_2}=\dfrac{4,8}{32}=0,15(mol)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,PTHH:4M+xO_2\xrightarrow{t^o}2M_2O_x\\ 2M+2xHCl\to 2MCl_x+xH_2\\ \Rightarrow \Sigma n_{M}=\dfrac{0,6}{x}+\dfrac{0,3}{x}=\dfrac{0,9}{x}\\ \Rightarrow M_{M}=\dfrac{8,1}{\dfrac{0,9}{x}}=9x(g/mol)\\ \text {Thay }x=3 \Rightarrow M_{M}=27(g/mol)\\ \text {Vậy M là nhôm (Al)}\)
\(b,\text {Dung dịch B là }AlCl_3\\ n_{Al}=\dfrac{8,1}{27}=0,3(mol)\\ \Rightarrow n_{AlCl_3}=n_{Al}=0,3(mol)\\ n_{Al(OH)_3}=\dfrac{15,6}{78}=0,2(mol)\\ PTHH:3NaOH+AlCl_3\to Al(OH)_3\downarrow +3NaCl\\ \text {Vì }\dfrac{n_{AlCl_3}}{1}>\dfrac{n_{Al(OH)_3}}{1} \text {nên } AlCl_3 \text { dư}\\ \Rightarrow n_{NaOH}=3n_{Al(OH)_3}=0,6(mol)\\ \Rightarrow V_{dd_{NaOH}}=0,6.2=1,2(l)\)
\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{5,28}{24}=0,22mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,22...0,44........0,22........0,22\\ V_{H_2\left(đkc\right)}=0,22.24,79=5,4538l\\ b)C_{\%HCl}=\dfrac{0,44.26,5}{200}\cdot100=8,03\%\\ c)C_{\%MgCl_2}=\dfrac{0,22.95}{200+5,28}\cdot100\approx10,18\%\)
a) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,3--->0,6------>0,3-->0,3
=> V = 0,3.24,79 = 7,437 (l)
b) a = mHCl = 0,6.36,5 = 21,9 (g)
b = mZnCl2 = 0,3.136 = 40,8 (g)
\(n_{HCl}=\dfrac{10\%.109,5}{36,5}=0,3\left(mol\right);n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.0,1=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,1=2,479\left(l\right)\\ b,ddA:HCl\left(dư\right),MgCl_2\\ m_{ddA}=2,4+109,5-0,1.2=111,7\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{111,7}.100\%\approx3,268\%;C\%_{ddMgCl_2}=\dfrac{0,1.95}{111,7}.100\%\approx8,505\%\)