How many grams of silver nitrate are required to produce 30 grams of silver? (Note: the equation is already balanced) Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)

Answer 1

Hi Emma!

When you are given grams of a product or reactant and then asked for grams of a different product or reactant, you will be doing a stoichiometry problem!

In stoichiometry, the basic pattern is this:

grams ----------> moles --------> moles ---------------> grams

molar mass ratio molar mass

STEP ONE: Write a chemical equation if one is not given.

STEP TWO: Balance the chemical equation if it is not already.

STEP THREE:

Start with the given in the upper left corner of the railroad track.

30 g Ag |

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|

STEP FOUR: We need to cancel out grams and convert to moles using the molar mass. To cancel grams, grams will have to go on the bottom. Moles will go on the top.

30 g Ag | mol Ag

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| g Ag

STEP FIVE: We need the molar mass to go by grams. The molar mass will always be for 1 mol.

The molar mass of Ag is 107.87 g/mol

30 g Ag | 1 mol Ag

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| 107.87 g Ag

STEP SIX: To "switch" substances, we need to use a mole:mole ratio from the balanced chemical equation.

Look back at the question to see what substance we need to convert to, which is AgNO3, and the substance we are starting with is Ag. This means that we need the coefficients of Ag and AgNO3 from the equation. In the equation, there are 2 mol of AgNO3 producing 2 mol Ag, so the ratio is 2:2.

To cancel mol of Ag, mol of Ag must go down and diagonally on the bottom. Mol of AgNO3 will go at the top.

30 g Ag | 1 mol Ag | 2 mol AgNO3

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| 107.87 g Ag | 2 mol Ag

STEP SEVEN: Convert back into grams by using molar mass (if the question is asking for grams).

SUBSTEP A: We need to calculate the molar mass of AgNO3. We will have to find the molar mass by adding up the mass of Ag, N, and Ox3 (there are 3 O in AgNO3, so the molar mass of O will have to be multiplied by 3).

Ag = 107.87

N = 14.01

O = 16*3 = 48

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169.88 g/mol

To cancel out mol of AgNO3, I will put 1 mol of AgNO3 on the bottom. We want grams, so grams will go on the top. *The molar mass always goes with grams, so the 169.88 g will go on the top.*

30 g Ag | 1 mol Ag | 2 mol AgNO3 | 169.88 g

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| 107.87 g Ag | 2 mol Ag | 1 mol AgNO3

STEP EIGHT: Multiply everything on top and divide by everything on the bottom.

I would enter it into the calculator like this:

30*2*169.88/107.87/2 = 47.24 g

STEP NINE: Apply appropriate significant figures.

The given from the question only had 1 significant figure, but I find it highly unlikely that your teacher would want you to round your answer to 1 sig fig, so let's just round to two. That would mean the final answer is 47 g.

If you would like another example of how to solve a stoichiometry problem that I answered, please view the following link: https://www.wyzant.com/resources/answers/892981/help-with-highschool-stoich-problem-please