Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,n_{CaCO_3}=\dfrac{200}{100}=2\left(mol\right)\\ Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
2 2
\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
\(\dfrac{1}{6}\) 2 \(\dfrac{2}{3}\) 2
\(n_{Fe\left(thu.được\right)}=\dfrac{266}{56}=4,75\left(mol\right)\)
\(\rightarrow n_{Fe\left(H_2\right)}=4,75-\dfrac{2}{3}=\dfrac{49}{12}\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(\dfrac{49}{24}\) 6,125 \(\dfrac{49}{12}\)
\(\rightarrow\left\{{}\begin{matrix}V_{CO}=2.22,4=44,8\left(l\right)\\V_{H_2}=6,125.22,4=137,2\left(l\right)\\m_{Fe_2O_3}=\left(\dfrac{1}{6}+\dfrac{49}{24}\right).160=\dfrac{1060}{3}\left(g\right)\end{matrix}\right.\)
TL:
Tham khảo nhé:
@@@@@@@@@@@@@@@@@@@@
@tuantuthan
HT
a) $CaCO_3 \xrightarrow{t^o} CaO + CO_2$
b) $m_{CaCO_3} = 120 - 120.20\% = 96(gam)$
Theo PTHH :
$n_{CaO} = n_{CaCO_3} = \dfrac{96}{100} = 0,96(mol)$
$\Rightarrow m_{CaO} = 0,96.56 = 53,76(gam)$
c) $n_{CO_2} = n_{CaCO_3} = 0,96(mol)$
$\Rightarrow V_{CO_2} = 0,96.22,4 = 21,504(lít)$
a. PTHH: 2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O
b. Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
Ta thấy: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\)
Vậy \(Al\left(OH\right)_3\) dư.
\(m_{dư}=0,75.78-98.0,5=9,5\left(g\right)\)
c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,5=\dfrac{1}{6}\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)
a, \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right);n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Mol: \(\dfrac{1}{3}\) 0,5 \(\dfrac{1}{6}\)
b, Ta có: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\) ⇒ Al(OH)3 dư, H2SO4 hết
⇒ \(m_{Al\left(OH\right)_3}=\left(0,75-\dfrac{1}{3}\right).78=32,5\left(g\right)\)
c, \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)
Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)
a. PTHH: 2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O
b. Không có chất dư (hoặc có thể bn cho sai 49(g) dung dịch là 49(g) H2SO4)
c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,75=0,375\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,375.342=128,25\left(g\right)\)
a. PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
b. Có: \(n_{CO_2}=\frac{11,2}{22,4}=0,5mol\)
\(\rightarrow n_{CaCO_3}=0,5mol\)
\(\rightarrow m_{CaCO_3}=0,5.100=50g\)
c. Có: \(n_{CO_2}=0,5mol\)
\(\rightarrow m_{CO_2}=0,5.44=22g\)