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Chemistry 15 Online
OpenStudy (anonymous):

How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure? Show all of the work used to find your answer. Ca + Cl2 CaCl2

OpenStudy (anonymous):

n=m/M n(Cl2)=n(Ca) V=n*Vm

OpenStudy (anonymous):

i got 31,29 dm3

sam (.sam.):

\[56gCa \times \frac{1molCa}{40gCa}\times \frac{1molCl_2}{1molCa}\times \frac{22.4dm^3Cl_2}{1molCl_2}=31.36dm^3Cl_2\]

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