Answered

If 18g of magnesium reacts with 33g of silver nitrate determine the limiting reactant​.

Sagot :

Angelalacson25viz

Answer:

Mg + 2AgNO3…..>Mg(NO3)2 +2Ag

Silver nitrate=216+14+48=278+14+48=340

No of moles of silver nitrate=33/340=0.097

No of moles of magnesium=0.049

Mass of magnesium needed 0.049×24=1.16 grams

Excess reactant is magnesium by 18–1.16=16.8g

Mass of silver formed =0.097×108=10.48 g