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CO(g) + 2 H2 --> CH3OH2.50 g of hydrogen is reacted with 30.0 L of carbon monoxide at STP. What is the volume of excess remaining in liters?

User TheG
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1 Answer

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The first step to solve this question is to convert the mass of H2 to moles, using its molecular mass:


2.50gH_2\cdot(molH_2)/(2.00gH_2)=1.25molH_2

Now, convert the volume of CO to moles using the ideal gas law:


PV=nRT

Solve the equation for n and replace P, V, T and R for its known values (1atm, 30.0L, 0.082atmL/molK and 273.15K):


\begin{gathered} n=(PV)/(RT) \\ n=(1atm\cdot30.0L)/(0.082atmL/molK\cdot273.15K) \\ n=1.34mol \end{gathered}

From the given equation we can find how many moles of CO react with 1.25 moles of H2:


1.25molH_2\cdot(1molCO)/(2molH_2)=0.625molCO

Substract this amount to the value of n to find the excess remaining:


1.34molCO-0.625molCO=0.715molCO

Use the ideal gas law once again, but this time to find the volume occupied by 0.715moles of CO:


\begin{gathered} V=(nRT)/(P) \\ V=(0.715mol\cdot0.082atmL/molK\cdot273.15K)/(1atm) \\ V=16.01L \end{gathered}

It means that 16.01L of CO are remaining.

User Selthien
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