Glucose + 2X -enzymes-> 2 Pyruvic acid + 4ATP. What substances is indicated by "X" in this equation?

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Glucose + 2X -enzymes-> 2 Pyruvic acid + 4ATP. What substances is indicated by "X" in this equation?

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Answer 1 · 2017-02-09T12:42:07

See explanation.

Explanation:

$Glucose + 2 X - enzymes \to 2 Pyruvic acid + 4 ATP$ $"Glucose" + 2"X" - "enzymes" -> 2 "Pyruvic acid" + 4"ATP"$

Let us represent the chemical reaction

$C_{6} H_{12} O_{6} + 2 X \to 2 C_{3} H_{4} O_{3} + 4 C_{10} H_{16} N_{5} O_{13} P_{3}$ $C_6H_12O_6 + 2X -> 2C_3H_4O_3 + 4C_10H_16N_5O_13P_3$

Now multiply the number of moles to the number of moles of each element

for example = $2 H_{2} = H 2 \cdot 2 = H 4$ $2H_2 = H2 * 2 = H4$

This is not true but helps in determining the missing reactant or missing product as in this problem.

The result would be

$C_{6} H_{12} O_{6} + 2 X \to C_{6} H_{8} O_{6} + C_{40} H_{56} N_{20} O_{52} P_{12}$ $C_6H_12O_6 + 2X -> C_6H_8O_6 + C_40H_56N_20O_52P_12$

Now show the number of atoms of each element in each compound we know.

$C_{6} H_{12} O_{6} = H_{12}, O_{6}, C_{6}$ $C_6H_12O_6 = H_12, O_6, C_6$
$C_{6} H_{8} O_{6} = H_{8}, O_{6}, C_{6}$ $C_6H_8O_6 = H_8, O_6, C_6$
$C_{40} H_{56} N_{20} O_{52} P_{12} = H_{56}, O_{52}, C_{40}, P_{12}$ $C_40H_56N_20O_52P_12 =H_56, O_52, C_40, P_12$

Add the no. of elements in the products and subtract it from the elements in reactants. it will give

$H_{56} + H_{8} - H_{12} = 52 atoms of hydrogen$ $H_56 + H_8 -H_12 = "52 atoms of hydrogen"$

$O_{52} + O_{6} - O_{6} = 52 atoms of oxygen$ $O_52 + O_6-O_6 = "52 atoms of oxygen"$

$C_{40} + C_{6} - C_{6} = 40 atoms of carbon$ $C_40 + C_6 - C_6 = "40 atoms of carbon"$

$P_{12} - 0 = P_{12}$ $P_12 - 0 = P_12$

so the remaining atoms would be $2 X$ $2"X"$ as energy is neither created nor destroyed, so

$X = \frac{C_{40} H_{52} O_{52} P_{12}}{2} = C_{20} H_{26} O_{26} P_{6}$ $"X" = (C_40H_52O_52P_12)/2 = C_20H_26O_26P_6"$

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Glucose + 2X -enzymes-> 2 Pyruvic acid + 4ATP. What substances is indicated by "X" in this equation?

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