Assume that magnesium would act atom-for-atom the same as copper in this experiment. how many grams of magnesium would have been used in the reaction if one gram of silver were produced? the atomic mass of magnesium is 24.31 g/mol.

0.1127 grams of Mg is used to produce one gram of silver

Explanation:

Write a balanced equation

Mg(s) +2Ag + (aq) --> (aq) + 2Ag(s)

The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol

1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]

1 g Ag of silver is produced from 0.1127 grams of Mg

0.1127 grams of Mg is used to produce one gram of silver

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g