Homework chapter 4 × owLv2] Online teaching( e A Student ls Asked To St x gnment

Question

Homework chapter 4 × owLv2] Online teaching( e A Student ls Asked To St x gnment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionLocator; assignme Revlew Toplcs) Use the References to access important values if needed for thi sodium iodide. For the following reaction, 0.408 moles of chlorine gas are mixed with 0.551 moles of chlorine(g) + sodium iodide(s) sodium chloride(s) + iodine(s) What is the formula for the limiting reagent? What is the maximum amount of sodium chloride that can be produced? moles Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

The balanced equation is

Cl2 + 2 NaI ----> 2 NaCl + I2

From the balanced equation we can say that

1 mole of Cl2 requires 2 mole of NaI so

0.408 mole of Cl2 will require

= 0.408 mole of Cl2 *(2 mole of NaI / 1 mole of Cl2 )

= 0.816 mole of NaI

BUt we have 0.551 mole of NaI which is in less amount. so sodium iodide is the limiting reactant

The formula of limiting reagent is: NaI

from the balanced equation we can say that

2 mole of NaI produces 2 mole of NaCl so

0.551 mole of NaI will produce

= 0.551 mole of NaI *(2 mole of NaCl / 2 mole of NaI)

= 0.551 mole of NaCl

Therefore, the amount of NaCl produced would be 0.551 mole

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