a, \(CaO+H_2SO_4\rightarrow CaSO_4+H_2O\)

b, \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CaSO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaSO_4}=0,1.136=13,6\left(g\right)\)

c, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

a) \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(CaO+H_2SO_4\)\(\rightarrow CaSO_4+H_2O\)

b) Ta có \(n_{CaSO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\rightarrow m_{CaSO_4}=0,1\cdot136=13,6\left(g\right)\)

c) Ta có \(n_{H_2SO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\rightarrow m_{H_2SO_4}=0,1\cdot98=9,8\left(g\right)\)

Zn+2HCl-->ZnCl2+H2

a, nZn=\(\dfrac{19,5}{65}\)=0,3mol

nH2=nZn=0,3mol

VH2=0,3*22,4=6,72l

b, nZnCl2=nZn=0,3mol

mZnCl2=0,3*(65+35,5*2)=40,8g

c,nHCl=2nZn=0,6mol

mHCl=0,6*36,5=21,9g

C%HCl=\(\dfrac{21.9}{200}\)*100%=10,95%

a) pthh: Zn + 2Hcl = \(ZnCl_2\) + \(H_2\)

\(_{_{ }}\)\(N_{ZN}\) = \(\dfrac{m}{M}\) =\(\dfrac{19,5}{65}\) =0.3 mol 

\(N_{H_2}\)= \(N_{Zn}\) = 0.3 mol 

\(V_{H_2}\)= n × 24.79 = 0.3 × 24.79 = 7.437 ( L)

b) \(N_{ZnCl_2}\)= \(N_{Zn}\) = 0.3 mol 

\(m_{ZnCl_2}\)= n × M = 0.3 × 136 = 40.8 g

c) \(m_{dd}\) = 19.5 + 200 = 219.5 g

\(C\%\:=\dfrac{m_{Ct}}{m_{dd}}\) × 100 =\(\dfrac{19.5}{219.5}\)×100= 8.88 %

oxide: Na2O: sodium oxide

            ZnO: zinc oxide

               Fe2O3: iron(II) oxide

               N2O3: nitrogen oxide 

    acid: HNO3: hydrogen nitrate

              HCl :  hydro chloric acid

             H2S : hydro sulfua

                    H2SO4: hydrogen sulfate

base: Ba(OH)2 : barium hydroxide

          Al(OH)3: aluminium hydroxide

 muoi:   Al(NO3)3: aluminium nitrate

              CaCO3: calcium carbonate

\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

_______________0,15______0,05 (mol)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

Giúp mình với

- Acid: 

+ \(HCl\): hydrochloric acid

+ \(H_2SO_4\): sulfuric acid

+ \(CH_3COOH\): acetic acid

- Base:

+ \(Ca\left(OH\right)_2\): calcium hydroxide

+ \(KOH\): potassium hydroxide

- Oxide acid: \(SO_3\): sulfur trioxide

- Oxide base: \(FeO\): iron (II) oxide

- Muối: 

+ \(BaSO_4\): barium sulfate

+ \(NaCl\): sodium chloride

Ta có: \(\left\{{}\begin{matrix}n_{NaHCO_3}=0,1.1,5=0,15\left(mol\right)\\n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=0,25.1=0,25\left(mol\right)\)

Nhỏ hh NaHCO₃ và Na₂CO₃ vào HCl nên NaHCO₃ và Na₂CO₃ pư đồng thời với HCl.

PT: \(NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

Mà: n_HCl < n_NaHCO3 + 2n_Na2CO3 nên HCl pư hết.

Có: \(\dfrac{n_{NaHCO_3}}{n_{Na_2CO_3}}=\dfrac{0,15}{0,1}=\dfrac{3}{2}\)

Gọi: n_{NaHCO3 (pư)} = 3x (mol) → n_{Na2CO3 (pư)} = 2x (mol)

Theo PT: n_HCl = n_{NaHCO3 (pư)} + 2n_{Na2CO3 (pư)}

⇒ 0,25 = 3x + 2.2x

⇒ x = 1/28

Theo PT: n_CO2 = n_{NaHCO3 (pư)} + n_{Na2CO3 (pư)} = 3.1/28 + 2.1/28 = 5/28 (mol)

⇒ V_CO2 = 5/28.22,4= 4 (l)

a, Pư xảy ra: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

b, \(n_{CuSO_4}=0,25.0,5=0,125\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{Cu}=n_{CuSO_4}=0,125\left(mol\right)\)

⇒ m_{thanh kẽm giảm} = m_Zn - m_Cu = 0,125.65 - 0,125.64 = 0,125 (g)