Silver nitrate is reacted with excess copper (II) chloride, producing solid silver chloride precipitate. If 25.5g...
Silver nitrate is reacted with excess copper (II) chloride, producing solid silver chloride precipitate. If 25.5g of silver nitrate is reacted and there is a 77.0% yield of silver chloride, how much silver was initially present?
AgNO3 (s) + CuCl2 (aq) → AgCl (s) + Cu(NO3)2 (aq)
Homework Answers
Answer:-
This question is solved by using the simple concept of stoichiometry which involves the relationship between masses of reactants and products.
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Silver nitrate is reacted with excess copper (II) chloride,
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For the following reaction, 5.09 grams of iron(II) chloride are mixed with excess silver nitrate Assume that the percent yield of iron(II) nitrate is 93.4% Iron(II) chloride(aq) + silver nitrate(aq) iron(II) nitrate(aq) + silver chloride(s) What is the ideal yield of iron(II) nitrate? What is the actual yield of iron(II) nitrate? grams grams Submit ANSWER Retry Entire Group 2 more group attempts remaining The equation for this reaction is: FeCl2(aq) + 2 AgNO3(aq) - Fe(NO3)2(aq) + 2 AgCl(s)
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Question 20 (1 point) Solid silver can be produced by reacting silver (1) nitrate with copper according to the reaction below. How how many moles of silver can be produced if 17.882 g of Cu was placed in silver (1) nitrate solution (Assume the AgNO3 is in excess) Cu(s) + 2 AgNO3(aq) + Cu(NO3)2(aq) + 2 Ag(s) Your Answer: Answer units Question 21 (1 point) . . . . n l .ncc. that
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