If a solution containing 18.37 g18.37 g of mercury(II) perchlorate is allowed to react completely with a solution containing 5.102 g5.102 g of sodium dichromate, how many grams of solid precipitate will form? precipitate:     How many grams of the reactant in excess will remain after the reaction? excess reactant:

If a solution containing 83.347 g of mercury (II) acetate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will form?

O Chemical Reactions Solving for a reactant in solution 0/5 Bis One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a 200. mL sample of groundwater known to be contaminated with iron(II) chloride, which would react with silver nitrate solution like this: FeCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Fe(NO3)2(aq) The chemist adds 72.0 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 5.9 mg of silver chloride. Calculate the concentration of iron(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. mg X 5

15.50 g of NH4Cl reacts with an excess of AgNO3.  In the reaction 35.50 g AgCl is produced. What is the actual yield of AgCl? NH4Cl  +  AgNO3 -->  AgCl  +  NH4NO3