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In the reaction Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq), how many moles of hydrogen gas will be produced from 250.0 milliliters of a 3.0 M HCl in an excess of Mg? 0.75 moles 0.38 moles 3.0 moles 1.5 moles

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IDC22

0.38 moles cuz if u do the equation its closer to 0.375

Answer: 0.38 moles of hydrogen

Explanation: Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

\(\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}\)

Thus \(\text{no of moles}of HCl={3.0M}\times {0.25 L}=0.75\)

\(Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)\)

\(HCl\) is a limiting reagent as it limits the formation of products and Mg is an excess reagent.

As 2 mole of \(HCl\) gives= 1 mole of \(H_2\)

0.75 moles of \(HCl\) will give=\(\frac{1}{2}\times 0.75=0.38moles\) of \(H_2\)

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