a) \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)
Đổi: \(V_{H_2SO_4}=150ml=0,15l\)
\(n_{H_2SO_4}=\dfrac{C_M}{V}=\dfrac{1}{0,15}=6,667\left(mol\right)\)
\(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PTHH, \(n_{H_2SO_4}>n_{Na}\) nên \(Na\) hết
Theo PTHH, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\cdot0,2=0,1\left(mol\right)\)
\(m_{Na_2SO_4}=n\cdot M=0,1\cdot142=14,2\left(g\right)\)
b) Theo PTHH, \(n_{H_2}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\cdot0,2=0,1\left(mol\right)\)
\(V_{H_2}=\dfrac{n}{22,4}=\dfrac{0,1}{22,4}=0,00446\left(l\right)\)
PTHH: 2Na + H2SO4 -> Na2SO4 + H2
a, \(n_{Na}\) = \(\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=1.0,15=0,15\left(mol\right)\)
Lập tỉ số: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\)
Vậy sau phản ứng Na hết, H2SO4 dư
Theo PTHH: \(n_{Na_2SO_4}=2n_{Na}=0,4\left(mol\right)\)
\(m_{Na_2SO_4}=0,4.142=56,8\left(g\right)\)
b, Theo PTHH: \(n_{H_2}=2n_{Na}=0,4\left(mol\right)\)
\(V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\)
c, \(V_{Na_2SO_4}=0,4\cdot22,4=8,96\left(l\right)\)
\(C_{M_{Na_2SO_4}}=\dfrac{0,4}{8,96}\approx0,04\left(M\right)\)
P/S: Sai nhớ báo nhé
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=C.V=1.0,15=0,15\left(mol\right)\)
PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)
Theo PTHH ta có tỉ lệ:
\(\dfrac{0,2}{2}=0,1< \dfrac{0,15}{1}\) => \(H_2SO_4\) dư. Na hết => tính theo \(n_{Na}\)
a. Theo PT ta có: \(n_{Na_2SO_4}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
b. Theo PT ta có: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(V_{Na_2SO_4}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow CM_{Na_2SO_4}=\dfrac{n}{V}=\dfrac{0,1}{2,24}=0,04\left(M\right)\)
c) \(CM_{Na_2SO_4}=\dfrac{n}{V}=\dfrac{0,1}{0,15}\approx0,67\left(M\right)\)