Alex drips hydrochloric acid onto a 5.9 g piece of magnesium in a single displacement reaction. How many molecules of HCl are required? 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g) how do you do this?

2.96 × 10²³ molecules is required if Alex drips hydrochloric acid onto a 5.9 g piece of magnesium in a single displacement reaction

According to this question, hydrochloric acid (HCl) reacts with magnesium metal (Mg) to produce magnesium chloride and hydrogen has as follows:

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Where; Mg = 24g/mol

mole = 5.9/24

mole = 0.246mol of Mg

To convert mole to number of molecules, we multiply number of moles by 6.02 × 10²³ (Avogadro's no.)

Hence, number of molecules of HCl required is 0.492mol × 6.02 × 10²³

= 2.96 × 10²³ molecules of HClHCl

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i think it's c. i'm not sure but i think so.

did anyone actually do this? like if anyone has submitted this i rly need .