\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,2(mol)\\ a,m_{BaSO_4}=0,2.233=46,6(g)\\ b,V_{dd_{BaCl_2}}=\dfrac{0,2}{1,5}\approx 0,13(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2+0,13}\approx 1,21M\)

\(d,\) Dd sau p/ứ là HCl nên làm quỳ tím hóa đỏ

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ n_{BaCl_2}=n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\\ b,V_{\text{dd}BaCl_2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)\\ c,C_{M\text{dd}HCl}=\dfrac{0,4}{\dfrac{2}{15}+0,2}=1,2\left(M\right)\\ d,V\text{ì}.c\text{ó}.\text{dd}.HCl\Rightarrow Qu\text{ỳ}.ho\text{á}.\text{đ}\text{ỏ}\)

\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\)

\(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

a) \(n_{SO_2}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow V=2,24l\)

b) \(n_{H_2SO_4}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow C_M=\dfrac{0,1}{0,2}=0,5M\)

c) \(m_{Na_2SO_4}=0,1\cdot142=14,2g\)

Chị làm cụ thể câu c cho em được khong ạ, sao nNa2SO4 lại =0,1 mol thế chị

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

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

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)

a) nMgSO4=0,2.1=0,2(mol)

PTHH:MgSO4+2KOH->Mg(OH)2\(\downarrow\)+K2SO4

(mol) 0,2 ->0,4 ->0,2 ->0,2

mMg(OH)2=0,2.58=11,6(g)

b) C%ddKOH=\(\dfrac{0,4.56}{200}.100\%\)=11,2%

c) mdd(sau pứ)=200+0,2.12011,6=212,4(g)

C%ddK2SO4=\(\dfrac{0,2.174}{212,4}.100\%\)=16,38%

á chết câu c mk làm lộn

c) mdd(sauphảnứng)=200.1,5+200- 11,6=488,4(g)

C%ddK2SO4=\(\dfrac{0,2.174}{488,4}.100\%=7,12\%\)

a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

b. Đổi 100ml = 0,1 lít

Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

a. PTHH: AgNO₃ + HCl ---> AgCl↓ + HNO₃

b. Ta có: \(n_{AgNO_3}=\dfrac{42,5}{170}=0,25\left(mol\right)\)

Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,25\left(mol\right)\)

=> \(m_{AgCl}=0,25.143,5=35,875\left(g\right)\)

c. Theo PT: \(n_{HCl}=n_{AgCl}=0,25\left(mol\right)\)

Đổi 100ml = 0,1 lít

=> \(C_{M_{HCl}}=\dfrac{0,25}{0,1}=2,5M\)