PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

a) Ta có: \(n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,05mol\) \(\Rightarrow m_{ddHCl}=\dfrac{0,05\cdot36,5}{18\%}\approx10,14\left(g\right)\)

b) Theo PTHH: \(n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,025\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,025\cdot111=2,775\left(g\right)\\m_{CO_2}=0,025\cdot44=1,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Zn}+m_{ddHCl}-m_{CO_2}=11,54\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{2,775}{11,54}\cdot100\%\approx24,05\%\)

CaCO₃+2HCl→CaCl₂+CO₂↑ +H₂O

\(+n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

\(+n_{HCl}=2n_{CaCO_3}=0,05\left(mol\right)\)

\(+m_{HCl}=0,05.98=4,9\left(gam\right)\)

\(+m_{dungdịchHCl}=\dfrac{4,9}{18}.100\%=27,2\left(gam\right)\)

\(+n_{CaCl}=n_{CaCO_3}=0,025\left(mol\right)\)

\(+m_{CaCl_2}=0,025.111=2,775\left(gam\right)\)

Theo ĐLBTKL ta có:

\(m_{CaCl_2}=2,5+27,2-0,025.44-0,025.18=28,15\left(gam\right)\)

C%=\(\dfrac{2,775}{28,15}.100\%\approx9,85\%\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ 0,05.........0,1..........0,05..........0,05\left(mol\right)\\ a.C\%_{ddHCl}=\dfrac{0,1.36,5}{200}.100=1,825\%\\ b.m_{Zn}=0,05.65=3,25\left(g\right)\\ c.C\%_{ddZnCl_2}=\dfrac{136.0,05}{3,25+200-0,05.2}.100\approx3,347\%\)

\(a,PTHH:FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe\left(OH\right)_2}=n_{FeCl_3}=\dfrac{10,7}{107}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_3}}=0,1\cdot162,5=16,25\left(g\right)\\ \Rightarrow m_{dd_{FeCl_3}}=\dfrac{16,25\cdot100\%}{5\%}=325\left(g\right)\\ b,n_{NaOH}=n_{NaCl}=3n_{Fe\left(OH\right)_3}=0,3\left(mol\right)\\ \Rightarrow m_{NaOH}=0,3\cdot40=12\left(g\right)\\ m_{NaCl}=0,3\cdot58,5=17,55\left(g\right)\\ \Rightarrow m_{dd_{NaCl}}=325+150-10,7=464,3\left(g\right)\\ \Rightarrow C\%_{dd_{NaCl}}=\dfrac{17,55}{464,3}\cdot100\%\approx3,78\%\)

Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

a. PTHH: Fe₃O₄ + 4H₂SO₄ ---> FeSO₄ + Fe₂(SO₄)₃ + 4H₂O

Theo PT: \(n_{H_2SO_4}=4.n_{Fe_3O_4}=4.0,01=0,04\left(mol\right)\)

=> \(m_{H_2SO_4}=0,04.98=3,92\left(g\right)\)

Theo đề, ta có: \(C_{\%_{H_2SO_4}}=\dfrac{3,92}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=19,6\left(g\right)\)

b. Ta có: \(m_{dd_{SauPỨ}}=2,32+19,6=21,92\left(g\right)\)

Theo PT: \(n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,01\left(mol\right)\)

=> \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,01.400=4\left(g\right)\)

=> \(m_{SauPỨ}=1,52+4=5,52\left(g\right)\)

=> \(C_{\%_{SauPỨ}}=\dfrac{5,52}{21,92}.100\%=25,18\%\)

PTHH: \(CH_3COOH+KHCO_3\rightarrow CH_3COOK+H_2O+CO_2\uparrow\)

a) Ta có: \(n_{CH_3COOH}=\dfrac{200\cdot24\%}{60}=0,8\left(mol\right)=n_{KHCO_3}\)

\(\Rightarrow m_{ddKHCO_3}=\dfrac{0,8\cdot100}{16,8\%}\approx476.2\left(g\right)\)

b) Theo PTHH: \(n_{CH_3COOK}=0,8\left(mol\right)=n_{CO_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COOK}=0,8\cdot98=78,4\left(g\right)\\m_{CO_2}=0,8\cdot44=35,2\left(g\right)\end{matrix}\right.\)

 Mặt khác: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)

\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)

Gọi số mol HNO₃ đã phản ứng là x (mol)

\(\Rightarrow n_{H_2O}=\dfrac{1}{2}x\left(mol\right)\)

a, Bảo toàn khối lượng ta có:

 \(5,21+63x=16,01+9x\Rightarrow x=0,2\Rightarrow m_{ddHNO_3}=200\left(g\right)\)

b, Ta có: \(n_{Zn\left(NO_3\right)_2}=a\left(mol\right);n_{Al\left(NO_3\right)_3}=2a\left(mol\right)\)

Tăng giảm khối lượng ta có: 

\(a.213-a.189=18,38-16,01\Rightarrow a=0,09875\\ \Rightarrow\%m_{ZnO}=\dfrac{0,09875.81.100\%}{5,21}=???\)

Đáp án > hơn 100% nên em xem lại đề nhé

a)nMgO=6:40=0,15(mol)

Ta có PTHH:

MgO+->MgS

0,15......0,15...........0,15..................(mol)

Theo PTHH:mH2SO4=0,15.98=14,7g

b)Ta có:mddH2SO4=D.V=1,2.50=60(g)

=>Nồng độ % dd  là:

C%ddH2SO418\66.100%=27,27%

300ml = 0,3l

\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)

Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)

          1               1               1              1

         0,3            0,3            0,3

\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

200ml = 0,2l

\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)

Sau phản ứng :

\(V_{dd}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

 Chúc bạn học tốt

\(n_{HNO_3}=0,3\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)

\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)

Bài này anh có hỗ trợ ở dưới rồi nha em!

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) KL dung dịch H2SO4 tham gia p.ứ là đề bài cho 100(g)

d) mH2SO4=0,2.98=19,6(g)

=>C%ddH2SO4= (19,6/100).100=19,6%

a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$

b)

$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)

$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$

$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$