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balandron [24]
3 years ago
8

If 143.56 mL of 0.6653 M ammonium carbonate reacts with 175.37 mL of 0.8732 M chromium(III) sulfate in a double replacement reac

tion and produces 7.543 g of chromium(III) carbonate, what is the percent yield of the reaction
Chemistry
2 answers:
SSSSS [86.1K]3 years ago
8 0
I think it’s 7543 g of chromium
Cloud [144]3 years ago
4 0
ANSWER=83.42%

EXPLANATION:

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2 years ago
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
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CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

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<span>Cl- = 0.4 moles             (ANSWER)</span>

6 0
3 years ago
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