Question

In the replacement reaction of magnesium nitrate with tin, how many grams of magnesium would be produced if 58.9 mg of magnesium nitrate? Report your answer with 3 significant figures.

2 Mg(NO3)2 + Sn → Sn(NO3)4 + 2 Mg

Answer 1

The given reaction is:

2 Mg(NO₃)₂ + Sn → Sn(NO₃)₄ + 2 Mg

As per the stoichiometry of the reaction:

2 moles of magnesium nitrate produces 2 mole of magnesium

Or, 1mole of magnesium nitrate produces 1 mole of magnesium.

Now, Molar mass of Mg(NO₃)₂ = 24. 3 g/ mol (for Mg)+ 2 x (14 g/mol (for N)+ 3x16 g/mol (for O)

   = 148.3 g/mol

Moles of Mg(NO₃)₂=    (58.9 mg ) / (148.3 g/mol ) = 3.97 x10 ^-4 g

So, moles of Mg produced = moles of Mg(NO₃)₂ = 3.97 x10 ^-4

Mass of Magnesium produced = moles x molar mass of Mg  

=(3.97 x10 ^-4 g ) x (24.3 g/mol )

    =9.65 x 10^-3 g                The value has 3 significant digits (9, 6 and 5)