Solid zirconium dioxide, ZrO2, is reacted with chlorine gas in the presence of c

Question

Solid zirconium dioxide, ZrO2, is reacted with chlorine gas in the presence of carbon. The products of the reaction are ZrCl4 and two gases, CO2 and CO in the ratio 1:2. Write a balanced chemical equation for the reaction: ?ZrO2(s)+?Cl2(g)+?C(s)??ZrCl4(s)+?CO2(g)+?CO(g) Give your answer as an ordered set of numbers ?, ?, ?, ... Use the least possible integers for the coefficients. PtB Starting with a 55.9?g sample of ZrO2, calculate the mass of ZrCl4 formed, assuming that ZrO2, is the limiting reagent and assuming 100% yield.

Explanation / Answer

?ZrO2(s)+?Cl2(g)+?C(s)??ZrCl4(s)+?CO2(g)+?CO(g

for balance equation

(?,?,?,?,?,?) = (2,4,3,2,1,2)

Thus equation is

2ZrO2(s)+4Cl2(g)+3C(s)?2ZrCl4(s)+CO2(g)+2CO(g)

Mass of ZrO2 = 55.9g

moles of ZrO2 = 55.9/123.218 moles

Now in balanced equation we have 2 moles of ZrO2 producing same no of moles of ZrCl4

And also ZrO2 is limiting agent all molar conversions should be in relation with its moles

Thus mass of ZrCl4 =molar mass of ZrCl4*moles of ZrCl4=(233.04*55.9/123.218)(2/2)

= 105.722670389 g

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