1. Date:
[tex]m_{s_{HCl}} = 400g[/tex]
[tex]c_{HCl} = 35\%[/tex]
[tex]n_{H_2}=?[/tex]
[tex]A_H=1[/tex]
[tex]A_{Cl}=35,5[/tex]
Rezolvare:
[tex]c=\dfrac {m_d \cdot 100}{m_s} \implies m_d = \dfrac{c \cdot \:m_s}{100} [/tex]
[tex]m_{d_{HCl}} =\dfrac {c_{HCl} \cdot m_{s_{HCl}}} {100} = \dfrac{35 \cdot 4\not0 \not0g}{1 \not0 \not0} = \boxed{140g \: HCl}[/tex]
[tex]\mathcal M_{HCl}= A_{H} + A_{Cl} = 1 + 35,5 = 36,5 \dfrac{g}{mol} [/tex]
[tex]\mathcal M_{H_2} = 2 \cdot A_{H} = 2 \cdot 1= 2 \dfrac{g}{mol}[/tex]
[tex] \\ Zn+ \dfrac { 140g\over{2HCl}} { \small2 \cdot36,5g}→ZnCl_2+\dfrac { x\over { H_2}}{2g}[/tex]
[tex] \dfrac{140}{2 \cdot36,5} = \dfrac{x}{2} \implies x = \dfrac{140 \: \cdot \not2}{36,5 \: \cdot\not2} = \boxed {3,83g \: H_2}[/tex]
[tex]n=\dfrac {m} {\mathcal M} \implies n_{H_2}=\dfrac {m_{H_2}} {\mathcal M_{H_2}} = \frac{3,83 \not g}{2 \frac{ \not g}{mol}} = \boxed{ \boxed{ 1,91 \: mol \:H_2}}[/tex]
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