MnO2 + 4 HCI + Cl2 + MnCl2 + 2 H20

Given 145.7 grams of manganese (IV) oxide, how much hydrochloric acid (HCI) is
needed to use up the MnO2 completely?

Explanation:

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244.79g of HCl

Explanation:

MnO2 + 4HCI —> Cl2 + MnCl2 + 2H20

Molar Mass of MnO2 = 54.9 + (2x16) = 54.9 + 32 = 86.9g/mol

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 4 x 36.5 = 146g

From the equation,

86.9g of MnO2 required 146g of HCl.

Therefore, 145.7g of MnO2 will require = (145.7 x 146) / 86.9 = 244.79g of HCl

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