Determine the mass of magnesium (Mg) needed to react with 0.955 moles of silver nitrate (AgNO3).

Question

Determine the mass of magnesium (Mg) needed to react with 0.955 moles of silver nitrate (AgNO₃).

Answer 1

First we need to write out the balanced chemical formula:

Mg + 2 AgNO₃ ––> Mg(NO₃)₂ + 2 Ag

We can use the coefficients to determine the molar ratio between Mg and AgNO₃. For every 2 moles of AgNO₃ that react, 1 mole of Mg reacts. Now we can use cross-multiplication to solve for the moles of Mg.

1 molMg/2 molAgNO₃ = x molMg /0.955 molAgNO₃

x = (1 molMg)*(0.955 molAgNO₃)/(2 molAgNO₃) = 0.4775 mol Mg

Now we use the atomic mass (or molar weight) of Mg to convert moles into grams.

MW_Mg = 24.305 g/mol

(0.4775 mol)*(24.305 g/mol) = 11.61 g Mg

Round for significant figures and your final answer is:

11.6 g Mg