Final answer:
To react completely with 7.5 grams of AgNO3, 4.58 grams of BaCl2 are required based on the stoichiometry of the balanced chemical equation and the molar mass of the reactants.
Step-by-step explanation:
To determine how many grams of BaCl2 (barium chloride) are necessary to react with 7.5 g AgNO3 (silver nitrate), we must consider the stoichiometry of the reaction. The balanced chemical equation for the reaction between silver nitrate and barium chloride is:
2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)
From the balanced equation, we see that 1 mole of BaCl2 reacts with 2 moles of AgNO3. First, we calculate the number of moles of AgNO3 we have by using its molar mass (169.87 g/mol):
moles AgNO3 = (7.5 g) / (169.87 g/mol) ≈ 0.044 moles
Since the reaction requires 1 mole of BaCl2 for every 2 moles of AgNO3, we need 0.022 moles of BaCl2. Finally, we use the molar mass of BaCl2 (208.23 g/mol) to find the mass:
mass BaCl2 = 0.022 moles * 208.23 g/mol ≈ 4.58 g
Therefore, 4.58 grams of BaCl2 are necessary to completely react with 7.5 grams of AgNO3.