Answer:

Explanation:
The I₂ is the common substance in the two equations.
(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O
{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻
From Equation (1), the molar ratio of iodate to iodine is

From Equation (2), the molar ratio of iodine to thiosulfate is

Combining the two ratios, we get

This question is describing the following chemical reaction at equilibrium:

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

Thus, by recalling the Van't Hoff's equation, we can write:

Hence, we solve for the enthalpy change as follows:

Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
Learn more:
1) Balanced chemical equation:
2SO2 (g) + O2 (g) -> 2SO3 (l)
2) Molar ratios
2 mol SO2 : 1 mol O2 : 2 mol SO3
3) Convert 6.00 g O2 to moles
number of moles = mass in grams / molar mass
number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.
4) Use proportions with the molar ratios
=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2
=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.
5) Convert 0.375 mol SO2 to grams
mass in grams = number of moles * molar mass
molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol
=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g
Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.
1: True
2: True
3: False
4: False
(Question 2 might not be true, not sure)
Answer is: four iron atoms <span>are reacting with every three molecules of oxygen.
Balanced chemical reaction: 4Fe + 3O</span>₂ → 2Fe₂O₃.
From chemical reaction: n(Fe) : n(O₂) = 4 : 3.
Ratios of atoms of iron and molecules of oxygen is 4 : 3..
There is four atoms of iron and six atoms of oxygen on both side of reaction.