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4. Part of the SO_(2) That Is Introduced into the Atmosphere by Combustion of Sulfur Containing Compounds Endl up Being Converted to

Problemas

4. Part of the SO_(2) that is introduced into the atmosphere by combustion of sulfur containing compounds endl up being converted to sulfuric acid, H_(2)SO_(4) underline ( )SO_(2)+underline ( )O_(2)+underline ( )H_(2)Oarrow underline ( )H_(2)SO_(4) a) Determine the limiting reactant given that 250 mol SO_(2) is reacting with 24.0 mol O_(2) and 40.0 mol H_(2)O Using this information, what is the mass of H_(2)SO_(4) that can be formed? Answer: __ b) If 86gof H_(2)SO_(4) is produced, what were the amounts of each of the reactants? Answer: __

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Juan professionell · Tutor durante 6 años
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4.6 (159 Votos)

Respuesta

a) To determine the limiting reactant, we need to compare the mole ratio of the reactants with the stoichiometry of the balanced chemical equation.The balanced chemical equation for the conversion of to is: Given that 250 mol is reacting with 24.0 mol and 40.0 mol , we can calculate the mole ratio of the reactants:Mole ratio of to : Mole ratio of to : Comparing the mole ratios with the stoichiometry of the balanced equation, we can see that the mole ratio of to is greater than the stoichiometry, while the mole ratio of to is less than the stoichiometry. Therefore, is the limiting reactant.To calculate the mass of that can be formed, we can use the stoichiometry of the balanced equation: From the balanced equation, we can see that 2 mol reacts with 1 mol and 2 mol to produce 2 mol . Since is the limiting reactant, we can use its amount to calculate the amount of formed: The molar mass of is 98.09 g/mol, so the mass of that can be formed is: Therefore, the mass of that can be formed is 3923.6 g.b) If 86 g of is produced, we can use the stoichiometry of the balanced equation to calculate the amounts of each of the reactants.From the balanced equation, we can see that 2 mol reacts with 1 mol and 2 mol to produce 2 mol . The molar mass of is 98.09 g/mol, so 86 g of corresponds to: Since the stoichiometry of the balanced equation is 2 mol for every 2 mol , the amount of required is: Similarly, the amount of required is: Therefore, the amounts of each of the reactants required to produce 86 g of are 0.876 mol and 0.438 mol .