\(PTHH:2Al+H_2SO_4-^{t^o}>Al_2SO_4+H_2\uparrow\)

ap dung DLBTKL ta co

\(m_{Al}+m_{H_2SO_4}=m_{Al_2SO_4}+m_{H_2}\)

thay so ta co

\(5,4+m_{H_2SO_4}=34,2+0,6\\ =>m_{H_2SO_4}=34,2+0,6-5,4\\ =>m_{H_2SO_4}=29,4\left(g\right)\)

M(X)= 0,552 x 29= 16 (g/mol)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{33,75}{27}=1,25\left(mol\right)\)

PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)

Theo PTHH: \(n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.1,25=0,9375\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=n.22,4=0,9375.22,4=21\left(l\right)\)

Mà \(V_{O_2}=\dfrac{1}{5}V_{kk}\)

=> \(V_{kk}=5.V_{O_2}=5.21=84\left(l\right)\)

b) Theo PTHH: \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.1,25=0,625\left(mol\right)\)

=> \(m_{Al_2O_3}=n.M=0,625.102=63,75\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{O_2}=a\left(mol\right)\\n_{N_2}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a\left(g\right)\\m_{N_2}=28b\left(g\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{hh.khí}=n_{O_2}+n_{N_2}=a+b=0,8\\m_{hh.khí}=m_{O_2}+m_{N_2}=32a+28b=23,2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,6\left(mol\right)\end{matrix}\right.\left(TM\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a=32.0,2=6,4\left(g\right)\\m_{N_2}=m_{hh.khí}-m_{O_2}=23,2-6,4=16,8\left(g\right)\end{matrix}\right.\)

Mai kiểm tra 15' rồi hy vọng mn giúp

\(n_{N_2}=\dfrac{m}{M}=\dfrac{14}{28}=0,5\left(mol\right)\\ \Rightarrow V_{N_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\)

ok thanks

\(PTHH:MgCO_3-^{t^o}>MgO+CO_2\)

ap dung DLBTKL ta co

\(m_{MgCO_3}=m_{MgO}+m_{CO_2}\\ =>m_{MgO}=m_{MgCO_3}-m_{CO_2}\\ =>m_{MgO}=84-44\\ =>m_{MgO}=40\left(g\right)\)

a)

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{6,4}{32+16\cdot2}=0,1\left(mol\right)\\ V_{SO_2}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)

b)

\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ m_{CO_2}=n\cdot M=0,5\cdot\left(12+16\cdot2\right)=22\left(g\right)\)

a)
n_so2= \(\overline{\dfrac{m}{M}}\)=\(\dfrac{6,4}{64}\)=0,1 ( mol)
v_so2= n*22,4 = 0,1* 22,4=2,24(lít)

b) n_co2 = \(\dfrac{v}{24,79}\)= \(\dfrac{11,2}{22,4}\)= 0,5 (mol)

m_co2= n*m= 0,5 * 44= 22(gam)

xg r nè bn iu❤