Question

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.

Answer 1

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

Step-by-step explanation:

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

Answer 2

Answer: The percent yield of zinc carbonate is 5.91 %

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} ZnCO_3=(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=
`(1)/(1)* 1.7=1.7moles` of
`ZnCO_3`

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

percentage yield =
`\frac{\text {Experimental yield}}{\text {Theoretical yield}}* 100=(12.6g)/(213.2g)* 100=5.91\%`