Respuesta :
Answer:
6C(s) + 6H2(g) + 3O2(g) --------------------> C6H12O6(s).
Explanation:
One of the examples under the type of sugar; monosaccharide(simple sugar) is the glucose. Glucose(with IUPAC name as D-glucose) as a monosaccharide (simple sugar) has the molar mass of 180.156 g/mol and a Melting point of 146 °C.
Thus, the unbalanced chemical reaction for the formation of glucose is given below;
C(s) + H2(g) + O2(g) --------------------------> C6H12O6(s) ΔH°f = -1273.1KJ/mol.
The balanced chemical reaction for the formation of one mole of glucose is given as;
6C(s) + 6H2(g) + 3O2(g) --------------------> C6H12O6(s).
The balanced chemical equation for the standard formation reaction of solid glucose is: [tex]\mathbf{6C_{(s)} + 6H_{2(g)} +3O_2_{(g)} \to C_6H_{12}O_{6} _{(s)} \ \ \ }[/tex]
Glucose is a simple sugar of a class of carbohydrates known as monosaccharides. It is the principal metabolite for energy production in the human body.
The formation of solid glucose [tex]\mathbf{C_6 H_{12} O_6_{(s)}}[/tex] occurs by the reaction of carbon, hydrogen, and oxygen.
The reaction formation of the solid glucose is:
[tex]\mathbf{C_{(s)} + H_{2(g)} +O_2_{(g)} \to C_6H_{12}O_{6} _{(s)} \ \ \ \Delta H_f^0= -1273.1 \ kJ/mol}[/tex]
Now, the balanced chemical equation can be expressed as follows;
[tex]\mathbf{6C_{(s)} + 6H_{2(g)} +3O_2_{(g)} \to C_6H_{12}O_{6} _{(s)} \ \ \ }[/tex]
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