POR FAVOR NECESITO AYUDA CON ESTA TAREA DE QUÍMICA PARA EL LUNES QUIEN PUEDE AYUDARME LA TAREA DICE: RESUELVA LOS SIGUIENTES PROBLEMAS : VERIFIQUE SI LA SIGUIENTE ECUACIÓN CUMPLE CON LA LEY DE LA CONSERVACIÓN DE LA MASA... H CL + CA O.................... CA CL + H2 O
Mm HCl = 36.45 g/mol CaO = 56 g/mol CaCl2 =110.9 g/mol H2O = 18 g/mol
2 HCl + CaO → CaCl2 + H2O 2(36.45 g) + (56 g) = (110.9 g) + 18 g 72.9 g + 56 g = 110.9 g + 18 g 128.9 g = 128.9
al balancear la ecuacion se cumple la LEY DE LA CONSERVACION DE LA MASA
`````````````````````````````````````````````````````````````````````````````````````````` HCl + CaO → CaCl2 + H2O (36.45 g) + (56 g) = (110.9 g) + 18 g 36.45 g + 56 g = 110.9 g + 18 g 92.45 g = 128.9
SIN BALANCEAR NO SE CUMPLE LA LEY DE LA CONSERVACION DE LA MAS
Mm HCl = 36.45 g/mol
CaO = 56 g/mol
CaCl2 =110.9 g/mol
H2O = 18 g/mol
2 HCl + CaO → CaCl2 + H2O
2(36.45 g) + (56 g) = (110.9 g) + 18 g
72.9 g + 56 g = 110.9 g + 18 g
128.9 g = 128.9
al balancear la ecuacion se cumple la LEY DE LA CONSERVACION DE LA MASA
``````````````````````````````````````````````````````````````````````````````````````````
HCl + CaO → CaCl2 + H2O
(36.45 g) + (56 g) = (110.9 g) + 18 g
36.45 g + 56 g = 110.9 g + 18 g
92.45 g = 128.9
SIN BALANCEAR NO SE CUMPLE LA LEY DE LA CONSERVACION DE LA MAS