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? $Mg (s) + 2 HCl (a q) \to {MgCl}_{2} (a q) + H_{2} (g)$ $"Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g)$

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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? $Mg (s) + 2 HCl (a q) \to {MgCl}_{2} (a q) + H_{2} (g)$ $"Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g)$

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Answer 1 · 2016-05-07T04:30:01

$2.24 L$ $2.24 "L"$

Explanation:

Step 1: determine the reaction

The chemical equation is already given.

$Mg + 2 HCl \to {MgCl}_{2} + H_{2}$ $"Mg" + 2 "HCl" -> "MgCl"_2 + "H"_2$

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

Step 3: determine how much $H_{2}$ $"H"_2$ is produced.

$0.100 mol$ $0.100 "mol"$ of $Mg$ $"Mg"$ and $0.200 mol$ $0.200 "mol"$ of $HCl$ $"HCl"$ will produce $0.100 mol$ $0.100 "mol"$ of $H_{2}$ $"H"_2$ .

Step 4: determine the volume at STP

Each mole of ideal gas occupies $22.4 L$ $22.4 "L"$ at STP. Therefore, $0.100 mol$ $0.100 "mol"$ of $H_{2}$ $"H"_2$ will occupy $2.24 L$ $2.24 "L"$ of space.

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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)+2HCl(aq)→MgCl2(aq)+H2(g)"Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g)

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