Question #95958

Jul 3, 2015

You can see by the equation that every mole of silver nitrate "needs" half a mole of copper.

Explanation:

What we need to do is divide the grams of $A g N {O}_{3}$ by its molar mass to find the number of moles of $A g N {O}_{3}$, and then divide this by $2$.

${m}_{m} \left(A g N {O}_{3}\right) = 107.9 + 14.0 + 3 \cdot 16.0 = 169.9 g / m o l$

Example:
100 grams of $A g N {O}_{3}$ would be:

$\frac{100}{169.9} m o l \to \frac{1}{2} \cdot \frac{100}{169.9} m o l C u$ would be needed.

(or a net division by 339.8)