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Consider the following reaction:

\[ \text{SrBr}_2(aq) + \text{K}_2\text{SO}_4(aq) \rightarrow 2\text{KBr}(aq) + \text{SrSO}_4(s) \]

If 0.500 L of 2.50 M strontium bromide reacts with 0.250 L of 1.55 M potassium sulfate, what is the mass of the precipitate formed?

Answer :


To find the mass of precipitate formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction, thus limiting the amount of product that can be formed.

The given chemical reaction is:SrBr2(aq) + K2SO4(ag) → 2KBr(ag) + SrSO4(s)


To do this, we can use the stoichiometry of the reaction, which relates the amounts of reactants and products based on their respective coefficients.

Let's calculate the number of moles of SrBr2 and K2SO4:

Number of moles of SrBr2 = 2.50 mol/L * 0.500 L = 1.25 mol
Number of moles of K2SO4 = 1.55 mol/L * 0.250 L = 0.3875 mol

According to the balanced equation, 1 mole of SrBr2 reacts with 1 mole of K2SO4 to produce 1 mole of SrSO4. Therefore, the ratio of moles of SrBr2 to moles of K2SO4 is 1:1.

Since the number of moles of SrBr2 (1.25 mol) is greater than the number of moles of K2SO4 (0.3875 mol), K2SO4 is the limiting reactant.

Now, let's calculate the mass of SrSO4 precipitate formed. The molar mass of SrSO4 is 183.68 g/mol.


Mass of SrSO4 = 0.3875 mol * 183.68 g/mol = 71.24 g


Therefore, the mass of the precipitate, SrSO4, formed in the reaction is 71.24 grams.

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