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Mg + 2AgNO3 -> 2Ag + Mg(NO3)2

How many grams of magnesium (Mg) are needed to completely react with 3.00 moles of silver nitrate (AgNO3)?

A. 72.9g Mg
B. 255g Mg
C. 16.2g Mg
D. 36.5g Mg
E. 18.2g Mh

Sagot :

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Answer:

146 grams Mg

Explanation:

Mg + 2AgNO3 -> 2Ag + Mg(NO3)2

The balanced equation tells us we need 1 mole of Mg for every 2 moles of silver nitrate.

Since we have 3.00 moles of AgNO3, we'll need 2x that, or 6 moles of Mg.

The molar mass of Mg is 24.3 grams/mole

(6 moles Mg)(24.3 grams Mg/mole Mg) =  145.8 grams Mg.  146 grams with 3 sig figs.

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