MnO2 + 4 HCl Cl2 + MnCl2 + 2 H2O
2.)Given 145.7 grams of manganese (IV) oxide, how much hydrochloric acid (HCl) is needed to use up the MnO2 completely?
Mass of MnO2 = 145.7g
Amount of HCl =?
Moles of MnO2 = mass of MnO2/
molecular mass of MnO2
= 1.6749mol of MnO2
1 mol MnO2 = 4 mol HCl
1.6749 mol MnO2 = (4*1.6749) mol HCl
= 6.6997mol HCl
Mass of HCl = mol of HCl * molecular mass of HCl
= 6.6997* 36.5
= 244.5417g HCl
Thank you so much for your help.
my pleasure :)
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