cau 6 nha
a, ta co pthh
Zn+H2SO4\(\rightarrow\)ZnSO4+H2
2Al+3H2SO4\(\rightarrow\)Al2(SO4)3+3H2
Theo de bai ta co nH2=\(\dfrac{17,92}{22,4}=0,8mol\)
Goi x la so mol cua H2 tham gia vao pthh 1
So mol cua H2 tham gia vao pthh 2 la 0,8-x mol
Theo pthh1 ta co nZn=nH2=x mol
Theo pthh 2 ta co nAl=\(\dfrac{2}{3}nH2=\dfrac{2}{3}.\left(0,8-x\right)mol\)
Theo de bai ta co he pt
65.x+27.\(\dfrac{2}{3}.\left(0,8-x\right)\)=23,8
\(\Leftrightarrow\)65x+18.(0,8-x)=23,8
\(\Leftrightarrow\)65x +14,4-18x=23,8
\(\Leftrightarrow\)47x+14,4=23,8
\(\Leftrightarrow\)47x=23,8-14,4
\(\Leftrightarrow\)47x=9,4
\(\Rightarrow\)x=\(\dfrac{9,4}{47}=0,2\)mol
\(\Rightarrow\)so mol cua H2 tham gia vao pthh 2 la 0,8-0,2=0,6 mol
\(\Rightarrow\)so mol cua Zn la
nZn=nH2=0,2 mol
nAl=\(\dfrac{2}{3}nH2=\dfrac{2}{3}.0,6=0,4mol\)
\(\Rightarrow\)khoi lg cua Al = 0,4.27=10,8 g
khoi lg cua Zn = 0,2.65=13 g
\(\Rightarrow\)% khoi lg moi kim loai la
%mAl= \(\dfrac{10,8.100}{23,8}\approx45,4\%\)
%mZn= 100%-45,4%= 54,6%
b, Ta co Nong do dd H2SO4 = \(\dfrac{nH2SO4}{vH2SO4}\Rightarrow nH2SO4=2,5.\dfrac{100}{1000}=0,25mol\)
C4)
1)
\(2KClO_3-t^0\rightarrow2KCl+3O_2\\ O_2+P-t^0->P_2O_5\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ 4H_3PO_4+3Ca->4H_2+Ca_3\left(PO_4\right)_2+CaHPO_4\)
2)
\(O_2+2Cu-t^0\rightarrow2CuO\\ CuO+H_2->Cu+H_2O\\ H_2O+K_2O->2KOH\\ KOH+CO_2->H_2O+K_2CO_3\)
nếu không gấp thì chiều mai được không?
cau b mk chua hoc den nen chi biet lam nhu the thoi
