To solve this problem, we first need to write and balance the chemical equation for the reaction between strontium chloride and sulfuric acid:
SrCl2 + H2SO4 → SrSO4 + 2HCl
According to the balanced chemical equation, one mole of strontium chloride reacts with one mole of sulfuric acid to produce one mole of strontium sulfate and two moles of hydrochloric acid.
Next, we need to calculate the number of moles of sulfuric acid we have:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
moles of H2SO4 = 300.0 g / 98.08 g/mol
moles of H2SO4 = 3.057 mol
Finally, we can use the stoichiometry of the balanced chemical equation to determine the number of moles of strontium chloride that will react with 3.057 moles of sulfuric acid:
moles of SrCl2 = moles of H2SO4
moles of SrCl2 = 3.057 mol
Now we can calculate the mass of strontium chloride using its molar mass:
mass of SrCl2 = moles of SrCl2 x molar mass of SrCl2
mass of SrCl2 = 3.057 mol x 158.53 g/mol
mass of SrCl2 = 485.1 g
Therefore, 485.1 grams of strontium chloride will react with 300.0 grams of sulfuric acid.
Step-by-step explanation:
To solve this problem, we use stoichiometry, which is a method that relates the amount of reactants and products in a chemical reaction based on their balanced chemical equation. In this case, we first write and balance the chemical equation for the reaction between strontium chloride and sulfuric acid. Then, we calculate the number of moles of sulfuric acid given its mass and molar mass. Next, we use the stoichiometry of the balanced chemical equation to determine the number of ontium chloride that will react with the given amount of sulfuric acid. Finally, we calculate the mass of strontium chloride using its molar mass and the calculated number of moles. By following these steps, we can determine the mass of strontium chloride that will react with 300.0 grams of sulfuric acid.