Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nCuO=16/80=0,2(mol)
a) PTHH: CuO + H2SO4 -> CuSO4 + H2O
0,2___________0,2_____0,2(mol)
b) mCuSO4=160.0,2=32(g)
c) mH2SO4=0,2.98=19,6(g)
=>C%ddH2SO4= (19,6/100).100=19,6%
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot\dfrac{10}{40}=0,125\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,125\cdot98}{10\%}=122,5\left(g\right)\\m_{Na_2SO_4}=0,125\cdot142=17,75\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{17,75}{10+122,5}\cdot100\%\approx13,4\%\)
\(n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\)
PTHH:
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,125 0,375 0,125 0,375
\(m_{ddH_2SO_4}=\dfrac{0,375.98.100}{25}=147\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,125.406}{20+147}\approx30,39\%\)
\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,1\cdot160=16\left(g\right)\\ b,n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{200}\cdot100\%=4,9\%\)
a)
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$n_{H_2SO_4} = n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$C\%_{H_2SO_4} = \dfrac{0,02.98}{100}.100\% = 1,96\%$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,04(mol)$
$m_{NaOH} = 0,04.40 = 1,6(gam)$
c)
$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
Cu dư nên $n_{SO_2} = \dfrac{1}{2}n_{H_2SO_4} = 0,05(mol)$
$V_{SO_2} = 0,05.22,4 = 1,12(lít)$
Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2mol\)
PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Theo PTHH: \(n_{H2}so_4=^nCuO=0,2mol\)
\(\rightarrow m_{H_2SO_4}=0,2.98=19,6g\)
\(+)^mddH_2SO_4=\dfrac{^mH_2SO_4}{C\%}.100=196g\)
Đến đây thì bn bt lm chx ạ?