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How many liters of hydrogen gas is produced from 3.712 g of magnesium with 104.2ml of 1.385 mol/L HCL (aq) at SATP? Please show your work and explain it. I really need help with this question, it's on one of my quizzes.

User Weyhan
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1 Answer

4 votes

Answer:


V=1.61L

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:


Mg+2HCl\rightarrow MgCl_2+H_2

Next, we compute the reacting moles of each reactants:


n_(Mg)=3.712gMg*(1molMg)/(24.305 gMg)=0.153molMg


n_(HCl)=1.385(molHCl)/(L)*0.1042L=0.144molHCl

Then, as magnesium and hydrohloric acid are in a 1:2 molar ratio 0.153 moles of magnesium will completely react with 0.306 moles of hydrochloric acid yet we only have 0.144 moles, therefore, limiting reactant is hydrochloric acid. Thus, we compute the produced moles of hydrogen:


n_(H_2)=0.144molHCl*(1molH_2)/(2molHCl) =0.072molH_2

Finally, we use the ideal gas equation with T=298K and 1atm (STP conditions) to compute the liters of hydrogen gas:


PV=nRT\\\\V=(nRT)/(P)=(0.072mol*0.082(atm*L)/(mol*K)*273K)/(1atm)\\ \\V=1.61L

Best regards.

User Tad Donaghe
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