a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

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

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

Của dd axit nha

PTHH: \(2AgNO_3+CaCl_2\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

Ta có: \(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{AgCl}=0,01\left(mol\right)\\n_{CaCl_2}=n_{Ca\left(NO_3\right)_2}=0,005\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,005\cdot111=0,555\left(g\right)\\m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,07+0,03}=0,05\left(M\right)\end{matrix}\right.\)

Ta có phương trình sau:

nAg = 2.16/108 = 0.02(mol)

C6H12O6 + Ag2O(NH3) => 2Ag + C6H12O7

0.01------------------------->0.02(mol)

Vì n(C6H12O6) = 1/2nAg = 1/2*0.02 = 0.01(mol)

=>CM(C6H12O6) = 0.01/0.05 = 0.2M

=> Đáp án A

a,2AgNO3+CaCl2=>Ca(NO3)2+2AgCl

b, nAgNO3=17/170=0,1(mol)

Theo PTPU, ta có nAgCl=nAgNO3=0,1(mol)

=>mAgCl=0,1.143,5=14,35(g)

c, ta có 2nCaCl2=nAgNO3=0,05(mol)

=>CM(CaCl2)=0,05/0,1=0,5(M)

đề có sai chỗ nào ko vậy

mik thấy thay Zn bằng Cu hợp lí hơn

$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : 

$n_{H_2} = n_{Zn} = \dfrac{12,8}{65} = 0,197(mol)$
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
Theo PTHH : 

$n_{H_2O} = n_{H_2} = 0,197(mol)$
$m_{H_2O} = 0,197.18 = 3,546(gam)$

\(n_{Zn}=\dfrac{12,8}{65}=\dfrac{64}{325}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ H_2+\dfrac{1}{2}O_2\underrightarrow{^{to}}H_2O\\ n_{H_2O}=n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{64}{325}\left(mol\right)\\ m_{H_2O}=\dfrac{64}{325}.18\approx3,545\left(g\right)\\ m_{ZnCl_2}=\dfrac{64}{325}.136\approx26,782\left(g\right)\)

\(a,n_{CaCl_2}=0,2\cdot0,1=0,02\left(mol\right)\\ n_{AgNO_3}=0,1\cdot0,1=0,01\left(mol\right)\\ PTHH:CaCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Ca\left(NO_3\right)_2\\ \text{Vì }\dfrac{n_{CaCl_2}}{1}>\dfrac{n_{AgNO_3}}{2}\Rightarrow CaCl_2\text{ dư}\\ \Rightarrow n_{AgCl}=0,01\left(mol\right)\\ \Rightarrow m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\ b,n_{Ca\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgCl}=0,005\left(mol\right)\\ \Rightarrow C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,1+0,1}=0,025M\)

Mọi người ơi giúp mik vs ạ . cảm ơn mn rất nhiều.

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

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