What are the mass and the identity of the precipitate that forms when 30.0 mL of 0.150 M HCl reacts with 25.0 mL of 0.200 M AgNO3?

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Write the balanced chemical equation for the reaction: \( \text{HCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{HNO}_3 \).

Calculate the moles of \( \text{HCl} \) using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Calculate the moles of \( \text{AgNO}_3 \) using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Determine the limiting reactant by comparing the mole ratio from the balanced equation with the calculated moles of \( \text{HCl} \) and \( \text{AgNO}_3 \).

Calculate the mass of the precipitate \( \text{AgCl} \) formed using the moles of the limiting reactant and the molar mass of \( \text{AgCl} \).

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Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble product, known as a precipitate. In this case, the reaction between hydrochloric acid (HCl) and silver nitrate (AgNO3) leads to the formation of silver chloride (AgCl), which is insoluble in water and precipitates out of the solution.

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Selective Precipitation

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced chemical equations. To determine the mass of the precipitate formed, one must first identify the limiting reactant by calculating the moles of each reactant and then use the stoichiometric ratios from the balanced equation to find the amount of product formed.

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Stoichiometry Concept

Molarity and Volume Calculations

Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. To find the number of moles of HCl and AgNO3, one can use the formula: moles = molarity × volume (in liters). This calculation is essential for determining how much of each reactant is available for the reaction and ultimately helps in identifying the mass of the precipitate formed.

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