C10H12O4S(s) + O2(g)  CO2(g) + SO2(g) + H2O(g)

26. When the equation above is balanced and all coefficients are reduced to their lowest whole-number terms, the coefficient for O2(g) is

(A) 6 (B) 7 (C) 12 (D) 14 (E) 28

The coefficient of O₂ when the equation is balanced and all the coefficients

are reduced to their lowest whole number term is 12.

The chemical equation is represented as follows:

C₁₀H₁₂O₄S(s) + O₂(g) → CO₂(g) + SO₂(g) + H₂O(g)

Now let's balance the equation and make sure the number of moles of

element in the reactant and product side are the same.

Therefore, the balance chemical equation will be as follows:

C₁₀H₁₂O₄S(s) +12 O₂(g) → 10CO₂(g) + SO₂(g) + 6H₂O(g)

From the balanced equation we can see that oxygen molecule has a

coefficient of 12 on the reactant side.

The number of moles of oxygen on both sides of the equation is 28.

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The coefficient for is 12.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 12 molecules of oxygen in balanced chemical equation. Thus the coefficient for is 12.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o