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What are the conversions needed to calculate the volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO3)2?
2Cu(NO3)2 (aq) + 4 KI (aq) ----> 2CuI (s) + 4KNO3 (aq)
Answer choices:
A) mL Cu(NO3)2 solution ---> mL ---> L KI solution
B) mL Cu(NO3)2 solution ---> L Cu(NO3)2 solution ---> grams Cu(NO3)2 ---> grams KI ---> L KI solution
C) mL Cu(NO3)2 solution ---> L Cu(NO3)2 solution ---> molecules Cu(NO3)2 ---> molecules KI ---> mol KI ---> L KI solution
D) mL Cu(NO3)2 solution ---> grams Cu(NO3)2 solution ---> mol Cu(NO3)2---> mol KI ---> L KI solution
E) mL Cu(NO3)2 solution ---> L Cu(NO3)2 solution ---> mol Cu(NO3)2 ---> mol KI ---> L KI solution
Sagot :
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