PLEASE HELP! DUE TODAY! How many gram of magnesium react with 120 ml H3PO4 (with molar concentration 2M)?

3.89 g Mg

Explanation:

2M = 2 mol/L

120 mL = 0.120 L

2 mol/L *0.120 L = 0.240 mol H3PO4

                               3Mg + 2H3PO4 > Mg3(PO4)2 + 3H2

from reaction         3 mol   2 mol

given                  0.240mol  x mol

x = 0.240*2/3 = 0.16 mol Mg

M(Mg) = 24.3 g/mol

24.3 g/mol *0.16 mol = 3.89 g Mg  

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%