Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to completion. How many liters of hydrogen gas, measured at STP, can be produced? $"Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g)$

Md Ariful Islam

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Step 1: determine the reaction

The chemical equation is already given.

$"Mg" + 2 "HCl" -> "MgCl"_2 + "H"_2$

Step 2: determine which reactant is limiting

0.100 mol of Mg reacts with 0.500 mol of HCl, which of them will run out first?

From the reaction, the ratio of $"Mg"$ to $"HCl"$ used is $1:2$. This means that if the reaction goes to completion, the $0.100 "mol"$ of $"Mg"$ will react with $0.100 "mol" xx 2 = 0.200 "mol"$ of $"HCl"$. This leaves $0.500 "mol" - 0.200 "mol" = 0.300 "mol"$ of $"HCl"$.

Step 3: determine how much $"H"_2$ is produced.

$0.100 "mol"$ of $"Mg"$ and $0.200 "mol"$ of $"HCl"$ will produce $0.100 "mol"$ of $"H"_2$.

Step 4: determine the volume at STP

Each mole of ideal gas occupies $22.4 "L"$ at STP. Therefore, $0.100 "mol"$ of $"H"_2$ will occupy $2.24 "L"$ of space.

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