#4HCl + MnO_2 -> 2H_2O + MnCl_2 + Cl_2#
For this reaction determine the mass of HCl that will react with
5.00 grams of manganese dioxide.
This is a basic stoichiometry problem of mass to mass conversions.
# "grams" -> "moles" -> "moles" -> "grams"#
Mass of #MnO_2 = 54.94 + 2(15.99) = "86.92 g MnO"_2#
Mass of #HCl = 1.01 + 35.45 = "36.46 g HCl"#
#5.00 cancel("g MnO"_2) xx (1 cancel("mol MnO"_2))/(86.92 cancel("g MnO"_2)) xx (4 cancel("mol HCl"))/(1cancel("mol MnO"_2)) xx "36.46 g HCl"/(1 cancel("mol HCl"))#
The middle conversion factor is the mole to mole ratio from the balance chemical equation of #"4 mol HCl"# for #"1 mol MnO"_2#
#((5.00) (4) (36.46))/(86.92)#
#"8.39 grams HCl"#
Here is a gram to gram video explanation.
http://bit.ly/1kljEVR
