Zinc metal reacts with silver nitrate according to the reaction:

zn(s) + 2agno3(aq)zn(no3 )2 (aq) + 2ag(s)

calculate the mass of ag that forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate?

2.38 grams of silver forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate.

Explanation:

To calculate the number of moles, we use the equation:

....(1)

The chemical equation is:

By stoichiometry of the reaction;

2 moles of silver nitrate reacts with 1 mole of zinc

Thus 0.0220 moles silver nitrate react with= moles of zinc

Thus silver nitrate will acts as limiting reagent and zinc acts as excess reagent.

2 moles of silver nitrate produces 2 mole of silver

Thus 0.0220 moles silver nitrate react with= moles of silver

Thus 2.38 grams of silver forms when 3.00g of zinc metal is placed in an aqueous solution containing 3.75g of silver nitrate.

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

it should be all of them because it asks if you included all of those in your answer

explanation: