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Answer :
Final answer:
The molar concentration of NaNO3 produced when 3.00 mL of 5.00 M AgNO3 reacts with 5.00 mL of 3.00 M Na2SO4 is calculated by stoichiometry and found to be 3.75 M.
Explanation:
To find the molar concentration of NaNO3 produced in the reaction between AgNO3 and Na2SO4, we need to perform a stoichiometric calculation based on the chemical equation:
AgNO3 + Na2SO4 → Ag2SO4 + 2NaNO3
We begin by calculating the moles of AgNO3 that reacted:
Moles of AgNO3 = volume (L) × molarity (M) = 0.003 L × 5.00 M = 0.015 mol
From the equation, for every 1 mole of AgNO3 that reacts, 2 moles of NaNO3 are produced. Therefore, we will have 0.015 mol × 2 = 0.03 mol of NaNO3.
Both solutions together have a final volume of 3.00 mL + 5.00 mL = 8.00 mL or 0.008 L. The concentration then is:
Concentration of NaNO3 = moles / volume = 0.03 mol / 0.008 L = 3.75 M
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