Hydrobromic acid dissolves solid iron according to the following reaction: Fe(s)+2HBr(aq)→FeBr2(aq)+H2(g)

What mass of HBr (in g) would you need to dissolve a 2.8-g pure iron bar on a padlock?
What mass of H2 would be produced by the complete reaction of the iron bar?

8.1g

0.1g

Explanation:

The reaction expression is given as:

      Fe   +   2HBr   →  FeBr₂    +    H₂

Mass of pure iron given  = 2.8g

A. Mass of HBr needed to dissolve a padlock of the mass;

To solve this problem, we need to use the mole concept.

 Convert mass of the known iron to the number of moles.

  Number of moles =  

  Molar mass  = 56g/mol  

   Number of moles of iron  =   = 0.05mole

 1 mole of Fe will react with 2 mole of HBr

 0.05mole of Fe will react with 0.05 x 2  = 0.1mole of HBr

 Mass of HBr  = number of moles x molar mass

   Molar mass of HBr  = 1 + 80   =  81g/mol

 Mass of HBr  = 0.1 x 81  = 8.1g

B. What mass of H2 would be produced by the complete reaction of the iron bar

    Since:

         1 mole of Fe will produce 2 mole of hydrogen gas

        0.05mole of Fe will produce 2 x 0.05mole  = 0.1mole of hydrogen gas

 Mass of hydrogen gas  = number of moles x molar mass

                                          = 0.1 x 1

                                          = 0.1g

d.) circle motion requires a force to give it a centripetal acceleration.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.