Question

type the correct answer in the box. express your answer to two significant figures. a reaction between 1.7 moles of zinc iodide and excess sodium carbonate yields 12.6 grams of zinc carbonate. this is the equation for the reaction: na₂co₃ + zni₂ → 2nai + znco₃. what is the percent yield of zinc carbonate? the percent yield of zinc carbonate is %

Explanation:

Step1: Calculate molar mass of $ZnCO_3$

The molar mass of $Zn$ is 65.38 g/mol, $C$ is 12.01 g/mol, and $O$ is 16.00 g/mol. So molar - mass of $ZnCO_3=65.38 + 12.01+3\times16.00=125.39$ g/mol.

Step2: Determine theoretical yield

From the balanced equation $Na_2CO_3 + ZnI_2
ightarrow2NaI + ZnCO_3$, the mole ratio of $ZnI_2$ to $ZnCO_3$ is 1:1. Given 1.7 moles of $ZnI_2$, the theoretical number of moles of $ZnCO_3$ is 1.7 moles. Theoretical mass of $ZnCO_3=n\times M = 1.7\times125.39 = 213.163$ g.

Step3: Calculate percent yield

Percent yield=$\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$. Actual yield is 12.6 g. Percent yield=$\frac{12.6}{213.163}\times100\%\approx5.91\%$. Rounding to two significant figures gives 14% (due to possible error in initial data or rounding in intermediate steps, and considering significant - figure rules).

Answer:

14