Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
mhh=\(\dfrac{3,36}{22,4}.64+\dfrac{2,8}{22,4}.28+\dfrac{6,72}{22,4}.2=13,7gam\)
=> ý A
![](https://rs.olm.vn/images/avt/0.png?1311)
$23)$
$n_{Cl_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$\Rightarrow m_{Cl_2}=0,2.71=14,2(g)$
$\to B$
$24)$
$Cu+2H_2SO_{4(đ)}\to CuSO_4+SO_4\uparrow+2H_2O$
Tỉ lệ: $1:2:1:1:2$
$\to C$
$25)CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$
$\Rightarrow m_{CaCO_3}=m_{CO_2}+m_{CaO}$
$\Rightarrow m_{CaO}<m_{CaCO_3}\Rightarrow m_{rắn}$ giảm
$\to A$
$26)$ Bảo toàn KL:
$m_X=m_{oxit}+m_{CO_2}$
$\Rightarrow m_{oxit}=31,8-15,4=16,4(g)$
$\to B$
$27)$
$PTHH:4FeS_2+11O_2\xrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow$
$\Rightarrow x:y=4:11$
$\to A$
$28)$
$n_{Fe}=\dfrac{140}{56}=2,5(mol)$
$Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\uparrow$
Theo PT: $n_{CO}=1,5.n_{Fe}=3,75(mol)$
$\Rightarrow V_{CO(đktc)}=3,75.22,4=84(lít)$
$\to $ Không đáp án nào đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1(đvC)=\dfrac{1}{12}.1,9926.10^{-23}=1,6605.10^{-24}(g)\\ \Rightarrow m_{Al}=27(đvC)=27.1,6605.10^{-24}\approx 4,48.10^{-23}(g)\)
Chọn C
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}n_{Cl_2}+n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\\dfrac{n_{Cl_2}}{n_{O_2}}=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Cl_2}=0,1\left(mol\right)\\n_{O_2}=0,3\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Cl_2}=0,1.71=7,1\left(g\right)\\m_{O_2}=0,3.32=9,6\left(g\right)\end{matrix}\right.\)
=> mhh = 7,1 + 9,6 = 16,7(g)
Đặt $n_{Cl_2}=x(mol)\Rightarrow n_{O_2}=3x(mol)$
Mà $n_{hh}=n_{Cl_2}+n_{O_2}=\dfrac{8,96}{22,4}=0,4$
$\Rightarrow x+3x=0,4\Rightarrow x=0,1$
$\Rightarrow m_{Cl_2}=0,1.71=7,1(g);m_{O_2}=3.0,1.32=9,6(g)$
$\Rightarrow m_{hh}=7,1+9,6=16,7(g)$
![](https://rs.olm.vn/images/avt/0.png?1311)
4Al + 3O2 ---> 2Al2O3
nAl2O3 = 10,2 / 102 = 0,1 ( mol )
=> nAl = 2.0,1 = 0,2 ( mol )
=> mAl = 0,2 . 27 = 5,4 g
=> D
\(n_{Cl_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow m_{Cl_2}=0,4.71=28,4\left(g\right)\\ \Rightarrow Ch\text{ọn}.C\)