Question

Type the correct answer in the box. Express your answer to two significant figures.

Answer 1

Answer: The percent yield of zinc carbonate is 5.91 %

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

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

$ZnI_2$ is the limiting reagent as it limits the formation of product and $Na_2CO_3$ is the excess reagent.  

$Na_2CO_3+ZnI_2\rightarrow 2NaI+ZnCO_3$  

According to stoichiometry :

1 mole of produce =  1 mole of

Thus 1.7 moles of $ZnI_2$ will produce=$\frac{1}{1}\times 1.7=1.7moles$  of $ZnCO_3$

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

percentage yield = $\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100=\frac{12.6g}{213.2g}\times 100=5.91\%$

Answer 2

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 :))