Consider the chemical equation for the combustion of sugar. _C6H12O6(s) + _O2(g) _CO2(g) + _H2O(l) Which sequence of coefficients should be placed in the blanks to balance this equation? 1, 6, 6, 6 6, 1, 6, 1 3, 3, 3, 6 1, 3, 3, 6

Respuesta :

I'll just solve it, but you should put each sequence on a separate line please.

C6H12O6 + O2 ===> CO2 + H2O    The easiest way to start is to balance the Cs. There are 6 on the left, so you must have 6 on the right.

C6H12O6(s) + O2(g) ===> 6CO2(g) + H2O(l) Next you have to balance out the hydrogens. There are 12 on the left and 2 on the right, so you have to multiply the Hs on the right by 6

C6H12O6(s) + O2(g) ===> 6CO2(g) + 6H2O(l). The Oxygens are the snake in this question. Begin with the right. There are 12 in 6CO2 (6 * 2) = 12 and 6 more on the water (6 * 1) = 6. The total is 12 + 6 = 18

So on the left you have to have a 9 so that 2 * 9 = 18 But 6 of them are already there is the C6H12O6(s) So you need 12 more from the oxygen. 6*12 = 12

C6H12O6(s) + 6O2(g) ===> 6CO2(g) + 6H2O(l)

And it is complete.

The answer should be A I think. But it doesn't look right. It should be 1 6 6 6


The balanced chemical equation for the combustion of sugar is

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

The sequence of coefficients that will balance the equation are: 1, 6, 6, 6

To know which option is correct, we shall balance the equation. This is illustrated below:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by writing 6 before CO₂ as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 6 before H₂O as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O

There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by writing 6 before O₂ as shown below:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Now, the equation is balanced.

The coefficients are: 1, 6, 6, 6

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