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a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
Bạn bổ sung thêm số liệu của khí thoát ra nhé.
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a. PTHH: \(Zn+H_2SO_4--->ZnSO_4+H_2\uparrow\left(1\right)\)
b. Theo PT(1): \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Zn}=65.0,3=19,5\left(g\right)\)
c. Theo PT(1): \(n_{H_2SO_4}=n_{Zn}=0,3\left(mol\right)\)
Đổi 300ml = 0,3 lít
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)
d. PTHH: \(2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\left(2\right)\)
Theo PT(2): \(n_{NaOH}=2.n_{H_2SO_4}=2.0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)
\(\Rightarrow m_{dd_{NaOH}}=\dfrac{24.100\%}{20\%}=120\left(g\right)\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)
b, mZn = 0,15.65 = 9,75 (g)
c, CM (H2SO4) = 0,15/0,05 = 3 M
d, mZnSO4 = 0,15.161 = 24,15 (g)
Bạn tham khảo nhé!
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,02<-----------0,01-------->0,01
=> VH2 = 0,01.22,4 = 0,224 (l)
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,02}{0,2}=0,1M\)
b)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,02------>0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,2}=0,1\left(l\right)=100\left(ml\right)\)
a) PTHH: Zn + H2SO4 -> ZnSO4 + H2
nH2= 0,15(mol)
=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)
b) mZn=0,15.65=9,75(g)
c) CMddH2SO4= 0,15/ 0,05=3(M)
d) mZnSO4= 161. 0,15=24,15(g)
\(a,Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,4}{0,4}=1\left(M\right)\\ c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\\ n_{CH_3COOK}=n_{CH_3COOH}=0,4\left(mol\right)\\ V_{ddCH_3COOK}=400+400=800\left(ml\right)=0,8\left(l\right)\\ C_{MddCH_3COOK}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
0,4 0,2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(b,C_{M_{CH_3COOH}}=\dfrac{0,4}{0,4}=1M\)
\(c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,4 0,4 0,4
\(C_{M_{CH_3COOK}}=\dfrac{0,4}{0,4}=1M\)