\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)

ZnO+H₂SO₄->ZnSO₄+H₂O

0,1-----0,1-------0,1-------0,1 mol

n _ZnO=\(\dfrac{8,1}{81}\)=0,1 mol

m _H2SO4 =39,2g =>n _H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol

=>H2SO4 , dư 0,3 mol

=>m _H2SO4=0,3.98=29,4g

=>C%_H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%

=>C% _ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)

Cái này bắt tính %m dd sau pư nên cần tính m dung dịch sau pư
ta có m_{dung dịch(sau pư)}= m_{chất tan}+ m_{dung môi} - m_{khí và rắn nếu có}

a) 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

b) \(n_{NaOH}=\dfrac{100.20}{100.40}=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

______0,5----->0,25------>0,25

=> m_H2SO4 = 0,25.98 = 24,5 (g)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) m_Na2SO4 = 0,25.142 = 35,5 (g)

m_{dd sau pư} = 100 + 125 = 225 (g)

=> \(C\%\left(Na_2SO_4\right)=\dfrac{35,5}{225}.100\%=15,778\%\)

Ta có: \(n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)

\(n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Dung dịch A gồm: CuSO₄ và H₂SO₄ dư

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

Đề có cho dữ kiện gì liên quan đến dd NaOH không bạn nhỉ?

\(n_{H_2SO_4}=\dfrac{200.19,6}{100.98}=0,4mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4\left(A\right)}=n_{CuO}=n_{H_2SO_4}=0,4mol\\ n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\\Rightarrow\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow CuSO_4.pư.không.hết\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

0,3mol        0,6mol                                        0,3mol

\(m_{ddB}=0,4.80+200+0,6.40-29,4=226,6g\\ C_{\%Na_2SO_4\left(B\right)}=\dfrac{0,3.142}{226,6}\cdot100=18,8\%\)

\(a)CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ b)n_{CuO}=\dfrac{4}{80}=0,05mol\\ n_{H_2SO_4}=\dfrac{100.20}{100.98}=\dfrac{10}{49}mol\\ \Rightarrow\dfrac{0,05}{1}< \dfrac{10:49}{1}\rightarrow H_2SO_4.dư\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05mol\\ C_{\%CuSO_4}=\dfrac{0,05.160}{100+4}\cdot100=7,69\%\\ C_{\%H_2SO_4}=\dfrac{\left(10:49-0,05\right)98}{100+4}\cdot100=14,52\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

a, PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

b, Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=3n_{Fe_2O_3}=0,3\left(mol\right)\end{matrix}\right.\)

⇒ n_{H2SO4 (dư)} = 0,4 - 0,3 = 0,1 (mol)

Ta có: m _{dd sau pư} = m_Fe2O3 + m _{dd H2SO4} = 16 + 200 = 216 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{216}.100\%\approx18,52\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{216}.100\%\approx4,54\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{3}\Rightarrow H_2SO_4dư\)

\(m_{\text{dung dịch sau phản ứng}}=16+200=216\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0.1\cdot400}{216}\cdot100\%=18.51\%\)

\(C\%_{H_2SO_4}=\dfrac{\left(0.4-0.2\right)\cdot98}{216}\cdot100\%=9.1\%\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right);n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2(SO_4)_3+3H_2O\)

Theo đề : 0,1.........0,4

Lập tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\)=> Sau phản ứng H2SO4 dư, Fe2O3 phản ứng hết

=> Dung dịch sau phản ứng gồm Fe2(SO4)3 và H2SO4 dư

m_{dd sau phản ứng} =16+200=216(g)

=>\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{216}=18,52\%\)

\(C\%_{H_2SO_4dư}=\dfrac{\left(0,4-0,1.3\right).98}{216}=0,45\%\)