The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

$\rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6$

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2$

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

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4 0

2 years ago

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

vazorg [7]

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %

4 0

3 years ago