Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g) According to the above reaction, determine the limiting reactant when 5.6 moles of

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Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this,  we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6  moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

what are the óptions. pléase list them