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Type the correct answer in the box. Express your answer to two significant figures.
A reaction between 1.7 moles of zinc Ipdide and excess sodium carbonate ylelds 12.6 grams of zinc carbonate. This is the equation for the
reaction:
Na2CO3 + Zniz - 2Nal +
ZnCoz.
What is the percent yield of zinc carbonate?
The percent yield of zinc carbonate is
5.91
1X.


Type The Correct Answer In The Box Express Your Answer To Two Significant Figures A Reaction Between 17 Moles Of Zinc Ipdide And Excess Sodium Carbonate Ylelds class=

Sagot :

Answer: The percent yield of zinc carbonate is 5.91 %

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

 [tex]\text{Moles of} ZnCO_3=\frac{12.6 g}{125.4g/mol}=0.100moles[/tex]

[tex]ZnI_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Na_2CO_3[/tex] is the excess reagent.  

[tex]Na_2CO_3+ZnI_2\rightarrow 2NaI+ZnCO_3[/tex]  

According to stoichiometry :

1 mole of produce =  1 mole of

Thus 1.7 moles of [tex]ZnI_2[/tex] will produce=[tex]\frac{1}{1}\times 1.7=1.7moles[/tex]  of [tex]ZnCO_3[/tex]

Theoretical yield of [tex]ZnCO_3=moles\times {\text {Molar mass}}=1.7moles\times 125.4g/mol=213.2g[/tex]

percentage yield = [tex]\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100=\frac{12.6g}{213.2g}\times 100=5.91\%[/tex]

Answer:

the other person was right EXCEPT it says in 2 significant numbers so the answer is 5.9

Explanation:

i hope this helps. have a wonderful day :))

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