mH₂SO₄ trong 100 gam dung dịch = 100.10% = 10 gam

mH₂SO₄ trong 150 gam dung dịch = 150.25% = 37,5 gam

m dung dịch sau khi trộn = 100 + 150 = 250 gam

C% = \(\dfrac{10+37,5}{250}.100\%\)= 19% , chọn D

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{14,8}.100\%\approx72,97\%\\\%m_{MgO}\approx27,03\%\end{matrix}\right.\)

b, Ta có: \(n_{MgO}=\dfrac{14,8-0,4.27}{40}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{H_2}+n_{MgO}=0,7\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,7.98=68,6\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{68,6}{10\%}=686\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\n_{MgSO_4}=n_{MgO}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 14,8 + 686 - 0,6.2 = 699,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,2.342}{699,6}.100\%\approx9,78\%\\C\%_{MgSO_4}=\dfrac{0,1.120}{699,6}.100\%\approx1,72\%\end{matrix}\right.\)

a)\(Mg+H2SO4-->MgSO4+H2\)

x----------------------------------------------x(mol)

\(Zn+H2SO4--->ZnSO4+H2\)

y------------------------------------------y(mol)

\(nH2=\frac{1,568}{22,4}=0,07\left(mol\right)\)

Theo bài ra ta có hpt

\(\left\{{}\begin{matrix}24x+65y=3,73\\x+y=0,07\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,05\end{matrix}\right.\)

\(m_{Mg}=0,02.24=0,48\left(g\right)\)

\(m_{Zn}=0,05.65=3,25\left(g\right)\)

b)\(n_{H2SO4}=n_{H2}=0,07\left(mol\right)\)

\(m_{H2SO4}=0,07.98=6,86\left(g\right)\)

\(m_{ddH2SO4}=\frac{6,86.100}{15}=45,33\left(g\right)\)

c)\(H2SO4+2KOH-->K2SO4+2H2O\)

\(n_{KOH}=0,2.0,1=0,02\left(mol\right)\)

\(n_{H2SO4}dư=\frac{1}{2}n_{KOH}=0,01\left(mol\right)\)

\(m_{H2SO4}dư=0,01.98=0,98\left(g\right)\)

\(m_{H2SO4bđ}=0,98+6,86=7,84\left(g\right)\)

\(m_{ddH2SO4bđ}=\frac{7,84.100}{15}=53,267\left(g\right)\)

d) dd X gồm H2SO4 dư, MgSO4 và ZnSO4

\(m_{MgSO4}=0,02.120=2,4\left(g\right)\)

\(m_{ZnSO4}=0,05.161=8,05\left(g\right)\)

\(m_{dd}\) sau pư =\(m_{KL}+m_{ddHClbđ}-m_{H2}=3,73+53,267-0,14\)

\(=56,877\left(g\right)\)

\(C\%_{H2SO4}dư=\frac{0,98}{56,877}.100\%=1,67\%\)

\(C\%_{MgSO4}=\frac{2,4}{56,877}.100\%=4,2\%\)

\(C\%_{ZnSO4}=\frac{8,05}{56,877}.100\%=14,15\%\)

\(n_{H_2SO_4}=\dfrac{150.9,8\%}{98}=0,15\left(mol\right)\\ H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+H_2O+CO_2\\ n_{Na_2CO_3}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{ddNa_2CO_3}=\dfrac{0,15.106}{10,6\%}=150\left(g\right)\\ n_{CO_2}=n_{H_2SO_4}=0,15\left(mol\right)\\ m_{ddsaupu}=150+150-0,15.44=293,4\left(g\right)\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,15\left(mol\right)\\ C\%_{Na_2SO_4}=\dfrac{0,15.142}{293,4}.100=7,26\%\)

\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\\ n_{H_2SO_4}=\dfrac{14,7}{98}=0,3\left(mol\right)\\ PTHH:H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\\ Mol:0,3\rightarrow0,3\rightarrow0,3\rightarrow0,3\)

\(m_{Na_2CO_3}=0,3.106=31,8\left(g\right)\\ m_{ddNa_2CO_3}=\dfrac{31,8}{10,6\%}=300\left(g\right)\\ m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\\ m_{CO_2}=0,3.44=13,2\left(g\right)\\ m_{dd}=150+300-13,2=436,8\left(g\right)\\ C\%_{Na_2SO_4}=\dfrac{42,6}{436,8}=9,75\%\)

mH2SO4 =mdd H2SO4.C% : 100% = 400.9,8% :100% = 39,2 (g)

=> nH2SO4 = mH2SO4 : MH2SO4 = 39,2: 98 = 0,4 (mol)

PTHH: H2SO4 + Na2CO3 ---> Na2SO4 + CO2 + H2O

0,4 ---->0,4 -----------> 0,4 -------> 0,4 (mol)

a) Theo PTHH: nNa2CO3 = nH2SO4 = 0,4 (mol)

=> mNa2CO3 = nNa2CO3. MNa2CO3 = 0,4.106 = 42,4 (g)

=> mdd Na2CO3 = mNa2CO3. 100% : C% = 42,4.100% : 10% = 424 (g)

b) Theo PTHH: nCO2 = nH2SO4 = 0,4 (mol)

=> VCO2(đktc) = 0,4.22,4 = 8,96 (lít)

c) Theo PTHH: nNa2SO4 = nH2SO4 = 0,4 (mol)

=> mNa2SO4 = nNa2SO4. MNa2SO4 = 0,4.142 = 56,8 (g)

mdd A = mdd H2SO4 + mdd Na2CO3 = 400 + 424 = 824 (g)

dd A chứa Na2SO4

=> C% Na2SO4 = (mNa2SO4 : mddA).100% = (56,8 : 824).100% = 6,89%

Thí nghiệm 1 : 

\(n_{Mg} = a(mol) ; n_{Fe} = b(mol) \\ \Rightarrow 24a + 56b = 15,6(1)\\ Mg + H_2SO_4 \to MgSO_4 + H_2\\ Fe + H_2SO_4 \to FeSO_4 + H_2\\ n_{H_2} =a + b = 0,3(2)\\ (1)(2) \Rightarrow a = 0,0375 ; b = 0,2625\)

Thí nghiệm 2 : Vì khối lượng thí nghiệm 1 bằng 3 lần khối lượng thí nghiệm 2 nên ở thì nghiệm 2 : \(n_{Mg} = \dfrac{0,0375}{3}=0,0125(mol)\\ n_{Fe} = \dfrac{0,2625}{3} = 0,0875(mol)\\ \text{Bảo toàn electron : }\\ n_{SO_2} = \dfrac{0,0125.2 + 0,0875.3}{2} = 0,14375(mol)\\ m_X - m_{SO_2} = 5,2 - 0,14375.64 = -4\)

Vậy khối lượng dung dịch Z giảm 4 gam so với H₂SO₄ ban đầu.

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?

\(n_{NaOH}=1.0,5=0,5(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,25(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,25.98}{9,8\%}=250(g)\)