a.PTHH:Zn+2HCl----->ZnCl₂+H₂

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{13}{65}=0,2\left(mol\right)\)

b.Theo PTHH:\(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48\left(l\right)\)

Theo PTHH:\(n_{HCl}=2n_{Zn}=2.0,2=0,4\left(mol\right)\)

\(m_{HCl}=n_{HCl}.M_{HCl}=0,4.36,5=14,6\left(g\right)\)

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\(n_S=\dfrac{m_S}{M_S}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Theo PTHH:\(n_{SO_2}=n_S=0,1\left(mol\right)\)

\(V_{SO_2}=n_{SO_2}.22,4=0,1.22,4=2,24\left(l\right)\)

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CTHH của X là C₂H₄O₂

lớp 1 mà giải phương trình

x-12=55

x=55+12

x=67

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a.PTHH:4Na+O₂----->2Na₂O

\(n_{Na}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)

Theo PTHH:\(n_{Na_2O}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

\(m_{Na_2O}=n_{Na_2O}.M_{Na_2O}=0,5.62=31\left(g\right)\)

PTHH:Zn+2HCl----->ZnCl₂+H₂

V_HCl=100ml=0,1(l)

a.\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PTHH:\(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)

b.Theo PTHH:\(n_{HCl}=2n_{Zn}=2.0,1=0,2\left(mol\right)\)

\(C_M=\dfrac{n}{V}=\dfrac{0,2}{0,1}=2\left(M\right)\)

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PTHH:Mg+2HCl----->MgCl₂+H₂

a.\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

Theo PTHH:\(n_{Mg}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

\(m_{Mg}=n_{Mg}.M_{Mg}=0,5.24=12\left(g\right)\)

b.Theo PTHH:\(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

\(m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,5.95=47,5\left(g\right)\)

c.Theo PTHH:\(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

\(m_{H_2}=n_{H_2}.M_{H_2}=0,5.2=1\left(g\right)\)

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