Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a/2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ b/n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{AlCl_3}=n_{Al}=0,2mol\\ m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\ c/higro\Rightarrow hydrogen\\ n_{H_2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\\ V_{H_2}=0,3.24,79=7,437\left(l\right)\\ d/n_{HCl}=\dfrac{0,2.6}{2}=0,6\left(mol\right)\\ V_{HCl}=\dfrac{0,6}{2}=0,3\left(l\right)\)
\(a.2Al+6HCl->2AlCl_3+3H_2\\ b.m_{AlCl_3}=\dfrac{1}{3}.0,1.0,6.133,5=2,67g\\ c.V_{H_2}=\dfrac{1}{2}.0,06.24,79=0,7437\left(L\right)\\ d.a=\dfrac{1}{3}.0,1.0,6.27=0,54g\)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0,05 0,15
\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)
\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
0,1<---------------0,1<-----0,15
\(\Rightarrow\left\{{}\begin{matrix}a,m_{Al}=0,1.27=2,7\left(g\right)\\b,m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\end{matrix}\right.\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)
b, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
c, \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{0,15}{1,5}=0,1\left(l\right)=100\left(ml\right)\)
\(a/Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\\ b/n_{H_2}=n_{FeCl_2}=0,1mol\\ m_{FeCl_2}=0,1.127=12,7\left(g\right)\\ c/V_{H_2}=0,1.22,4=2,24\left(l\right)\\ d/n_{HCl}=0,1.2=0,2\left(mol\right)\\ V_{HCl\left(pư\right)}=\dfrac{0,2}{2}=0,1\left(l\right)\)
\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Còn lại giống câu dưới nha
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
_____0,2_______0,2______0,2____0,2 (mol)
a, \(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(m_{MgSO_4}=0,2.120=24\left(g\right)\)
c, \(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b,Theo.\text{Đ}LBTKL:\\ m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\\ \Leftrightarrow5,4+29,4=m+0,6\\ \Leftrightarrow m=\left(5,4+29,4\right)-0,6=34,2\left(g\right)\)
\(a/2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)
\(b/30ml=0,03l\\ n_{H_2SO_4}=0,5.0,03=0,0015\left(mol\right)\\ n_{Al}=\dfrac{0,0015.2}{3}=0,001\left(mol\right)\\ m_{Al}=0,001.27=0,027\left(g\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,0015}{2}=0,00075\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,00075.342=0,2565\left(g\right)\)
\(c/n_{H_2}=\dfrac{0,0015.3}{3}=0,0015\left(mol\right)\\ V_{H_2}=0,0015.24,79=0,037185\left(l\right)\)
\(a.2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ b.n_{Al}=1,5.0,5.0,03=0,0375mol\\ m_{Al}=0,0375.27=1,0125g\\ m_{Al_2\left(SO_4\right)_3}=342\cdot\dfrac{1}{3}\cdot0,03\cdot0,5=1,71g\\V_{H_2}=24,79.0,5.0,03=0,37185L\)