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Chapter 4: Problem 51

How many milliliters of \(0.650 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) are needed to precipitate all the silver in \(415 \mathrm{mL}\) of \(0.186 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) as \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s}) ?\) \(2 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq}) $$

Short Answer

Expert verified
59.3 mL of a 0.650 M solution of K2CrO4 are needed to precipitate all the silver in the 415 mL of 0.186 M AgNO3 solution.

Step by step solution

01

Write down the given information

Record what you know. You know the molarity (M) and volume (in mL) of both the silver nitrate and potassium chromate solutions. Specifically, \(M_{AgNO_{3}} = 0.186 M\), \(V_{AgNO3} = 415 mL\), and \(M_{K2CrO_{4}} = 0.650 M\). The goal is to find \(V_{K2CrO_{4}}\).
02

Convert volumes to Liters

To work within the confines of the concentration unit Molarity, which is moles/Liter, it's necessary to convert the volumes of the solutions from milliliters to liters. To do this, you know that 1L = 1000mL. This gives us that \(V_{AgNO3}= 415 mL = 0.415 L\).
03

Calculate moles of AgNO3

To calculate the amount of moles of AgNO3 present in the solution, you'll need to use the equation for Molarity, which is \(M(molarity) = \frac{moles}{volume(L)}\). Therefore, you get \(n_{AgNO3} = M_{AgNO3} * V_{AgNO3} = 0.186 M * 0.415 L = 0.0771 mol\).
04

Use stoichiometric ratios to calculate moles of K2CrO4

From the balanced chemical reaction, we can see that the stoichiometric ratio between AgNO3 and K2CrO4 is 2:1, meaning that for every 2 moles of AgNO3, 1 mole of K2CrO4 is required. Therefore, the number of moles of K2CrO4 needed is \(n_{K2CrO4} = \frac{n_{AgNO3}}{2} = \frac{0.0771 mol}{2} = 0.03855 mol\).
05

Calculate the volume of K2CrO4 in Liters

With the calculated moles of K2CrO4 and the given Molarity, the volume can be calculated using the formula \(V = \frac{n}{M}\). Substituting the known variables in, you find that \(V_{K2CrO4} (L) = \frac{n_{K2CrO4}}{M_{K2CrO4}} = \frac{0.03855 mol}{0.650 M} = 0.0593 L\) , or 59.3 mL after converting liters back to mL.

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Most popular questions from this chapter

It is desired to produce as large a volume of \(1.25 \mathrm{M}\) urea \(\left[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{aq})\right]\) as possible from these three sources: \(345 \mathrm{mL}\) of \(1.29 \mathrm{M} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}, 485 \mathrm{mL}\) of \(0.653 \mathrm{M}\) \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) and \(835 \mathrm{mL}\) of \(0.775 \mathrm{M} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} .\) How can this be done? What is the maximum volume of this solution obtainable?

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

Nitric acid, \(\mathrm{HNO}_{3}\), can be manufactured from ammonia, \(\mathrm{NH}_{3}\), by using the three reactions shown below. $$\begin{aligned} &\text { Step 1: 4 NH }_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\\\ &\text { Step 2: } 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\\\ &\text { Step 3: } 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{aligned}$$ What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not recycled back into step 2.) (a) 1.33 mol; (b) 2.00 mol; (c) 2.67 mol; (d) 4.00 mol; (e) 6.00 mol.

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$
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