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MArishka [77]
3 years ago
7

Write a double replacement precipitation reaction A precipitate forms when aqueous solutions of nickel(II) chloride and silver(I

) nitrate are combined.
Chemistry
1 answer:
harkovskaia [24]3 years ago
3 0

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Double displacement reaction is defined as the reaction in which exchange of ions takes place.

AB+CD\rightarrow CB+AD

Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

When nickel (II) chloride reacts with silver (I) nitrate, it leads to the formation of white precipitate of silver chloride and an aqueous solution of nickel (II) nitrate.

The balanced chemical equation for the above reaction follows:

NiCl_2(aq.)+2AgNO_3(aq.)\rightarrow 2AgCl(s)+Ni(NO_3)_2(aq.)

By Stoichiometry of the reaction:

1 mole of aqueous solution of nickel (II) chloride reacts with 2 moles of aqueous solution of silver (I) nitrate to produce 2 moles of solid silver chloride and 1 moles of aqueous solution of nickel (II) nitrate

Hence, the balanced chemical equation is written above.

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3 years ago
When 4.15 grams of silver nitrate is reacted with 1.11 grams of iron(III) chloride, which best represents the amount of silver c
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Answer:

The mass of silver chloride produced = 2.202 g

Explanation:

Equation of the reaction is given below

3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)

molar mass of AgNO₃ = 170 g/mol

molar mass of FeCl₃ = 233.5 g/mol

molar mass of AgCl = 143.5 g/mol

3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate

4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃

1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃

mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1

therefore, FeCl₃ is the limiting reactant

0.0047 moles of FeCl₃ reacting will produce 0.0047 *  3 moles of AgCl = 0.0141 moles of AgCl

0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =

Therefore mass of silver chloride produced = 2.202 g

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3 years ago
Create a chemical equation for "Bubbling chlorine gas through a solution of sodium bromide gives elemental bromine and a solutio
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Answer:

Cl_{2} +2NaBr ---> 2NaCl + Br_{2}

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