Given that 25.0 mL of 0.100 M AgNO₃ reacts completely with 0.150 M k₂CrO₄,what volume of potassium chromate is required? AgNO₃(aq)+ k₂CrO₄(aq)→ Ag2CrO₄(s)+ KNO₃(aq)

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To solve this problem, we need to use the stoichiometry of the reaction. The balanced chemical equation is already given:$$ \text{2AgNO}_3(aq) + \text{K}_2\text{CrO}_4(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2\text{KNO}_3(aq) $$From the equation, 2 moles of AgNO₃ react with 1 mole of K₂CrO₄. First, calculate the moles of AgNO₃ used:$$ \text{Moles of AgNO}_3 = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{moles} $$Since it takes 2 moles of AgNO₃ to react with 1 mole of K₂CrO₄, the moles of K₂CrO₄ needed are half of the moles of AgNO₃:$$ \text{Moles of K}_2\text{CrO}_4 = \frac{0.0025}{2} = 0.00125 \, \text{moles} $$Finally, calculate the volume of 0.150 M K₂CrO₄ solution required to provide 0.00125 moles:$$ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.00125}{0.150} = 0.00833 \, \text{L} = 8.33 \, \text{mL} $$

Given That 25