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Resultados Actividad 2 H agua =15.2cm, 70.93ml H h2 =6cm, 28ml T2=18.5°C Actividad 3 m vaso =50.68g m 2 =50.96g m cu =0.28g Análisis de resultados-Manejo de datos Actividad 2 Patm = P H2 + Pv (h2O) +PHidrostática Donde Patm= 585mmHg =0.585mHg =77809.68Pa Pv(h2O) a temperatura de 18.5°C= 15.477mmHg =0.015477mHg =2058.5648Pa Presión hidrostática: pgh=1000(9.78)(0.152)=1486.56Pa =11.15mmHg Patm - Pv (h2O) - PHidrostática = P H2 P H2 =77809.68-2058.5648-1486.56 =47264.5552Pa En mmHg P H2 = 585mmHg-15.477mmHg-11.15mmHg=558.373mmHg En atm P H2 =0.734atm Obteniendo el número de moles N h2 = ( 0.734 atm )( 0.028 l ) ( 0.08205 )( 291.65 ) = 8.5884 x 10 4 moles Rendimiento experimental Calculo teórico Ecuación química Zn (s) + 2HCL (ac) H2 (g) + ZnCl 2(ac) Ecuación balanceada Utilizamos 0.10g de cinc metálico, obtengamos el número de moles de zn

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n zn = 0.10 g 65.38 g / mol = 1.5295 x 10 3 mol de Zn Para el HCl ( 37 100 )( 1.19 )( 1000 1 ) 36.461 = 12.0759 M n hcl = 12.076(0.5x10 -3 )= 6.038x10 -3 Reactivo limitante Para Zn 1.5295 x 10 3 1 = 1.5295 x 10 3 Para HCL 6.038 x 10 3 2 = 3.019 x 10 3 El Zn es el reactivo limitante De la ecuación obtenemos que 1 mol de Zn produce 1 mol de H 2 Entonces el número de moles de H 2 es 1.5295 x 10 3
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• Summer '14
• ANTonio
• Volumen, Hidrógeno, Cinc, alumínio, Cobre

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