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PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{NaOH}=\dfrac{14,8}{40}=0,37\left(mol\right)\)
a, Theo PT: \(n_{Na}=n_{NaOH}=0,37\left(mol\right)\)
\(\Rightarrow A_{Na}=0,37.6.10^{23}=2,22.10^{23}\) (nguyên tử)
\(m_{Na}=0,37.23=8,51\left(g\right)\)
b, \(n_{H_2}=\dfrac{1}{2}n_{NaOH}=0,185\left(mol\right)\)
\(\Rightarrow A_{H_2}=0,185.6.10^{23}=1,11.10^{23}\) (phân tử)
\(m_{H_2}=0,185.2=0,37\left(g\right)\)
c, \(V_{H_2}=0,185.24,79=4,58615\left(l\right)\)
PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
Ta có: \(n_{CaCO_3}=\dfrac{2,1.10^{24}}{6.10^{23}}=3,5\left(mol\right)\)
a, Theo PT: \(n_{CaO}=n_{CO_2}=n_{CaCO_3}=3,5\left(mol\right)\)
\(\Rightarrow A_{CaO}=A_{CO_2}=3,5.6.10^{23}=2,1.10^{24}\) (phân tử)
b, \(m_{CaO}=3,5.56=196\left(g\right)\)
\(m_{CO_2}=3,5.44=154\left(g\right)\)
c, \(V_{CO_2}=3,5.24,79=86,765\left(l\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(n_{H_2}=\dfrac{30,24}{24,79}\approx1,22\left(mol\right)\)
a, \(m_{H_2}=1,22.2=2,44\left(g\right)\)
b, Theo PT: \(n_{Na}=2n_{H_2}=2,44\left(mol\right)\)
\(\Rightarrow A_{Na}=2,44.6.10^{23}=14,64.10^{23}\) (nguyên tử)
\(m_{Na}=2,44.23=56,12\left(g\right)\)
c, \(n_{NaOH}=2n_{H_2}=2,44\left(mol\right)\)
\(\Rightarrow A_{NaOH}=2,44.6.10^{23}=14,64.10^{23}\) (phân tử)
\(\Rightarrow m_{NaOH}=2,44.40=97,6\left(g\right)\)
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Na}=0,3.23=6,9\left(g\right)\)
\(\Rightarrow m_{Na_2O}=13,1-6,9=6,2\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,5\left(mol\right)\)
Ta có: m dd sau pư = 13,1 + 200 - 0,15.2 = 212,8 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{212,8}.100\%\approx9,4\%\)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
bđ 0,3 0,4
pư 0,3 0,15
sau pư 0 0,25 0,3
=> H2 hết, O2 dư
\(m_{O_2\left(dư\right)}=0,25.32=8\left(g\right)\)
b) \(A_{H_2O}=0,3.6.10^{23}=1,8.10^{23}\left(phân.tử\right)\)
c) \(m_{O_2\left(pư\right)}=0,15.32=4,8\left(g\right)\)
PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,3<-------------------------------------0,15
\(\rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,2____________0,2____0,2 (mol)
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
c, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H2 dư.
Theo PT: \(n_{Cu}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{cr}=m_{Cu}=0,1.64=6,4\left(g\right)\)
\(a.2Na+2H_2O->2NaOH+H_2\\ n_{Na}=\dfrac{1,2\cdot10^{23}}{6\cdot10^{23}}=0,2mol=n_{NaOH}\\ n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ N_{NaOH}=N_{Na}=1,2\cdot10^{23}\left(PT\right)\\ N_{H_2}=0,1\cdot6\cdot10^{23}=0,6\cdot10^{23}\left(PT\right)\\ b.m_{NaOH}=40\cdot0,2=8g\\ m_{H_2}=2\cdot0,1=0,2g\\ c.V_{H_2}=22,4\cdot0,1=2,24L\)