Answer to Question #149255 in General Chemistry for riye

Question

When solutions of AgNO₃ and NaOH react, the balanced molecular equation is: 2 AgNO₃(aq) + 2 NaOH(aq) → Ag₂O(s) + 2 NaNO₃(aq) + H₂O(l) How much Ag₂O is produced when 0.200 g of AgNO₃ and 0.200 g of NaOH react?

Answer 1

Q149255

When solutions of AgNO₃ and NaOH react, the balanced molecular equation is:

2 AgNO₃(aq) + 2 NaOH(aq) → Ag₂O(s) + 2 NaNO₃(aq) + H₂O(l)

How much Ag₂O is produced when 0.200 g of AgNO₃ and 0.200 g of NaOH react?

Solution : The given problem is based on the limiting reactant.

Limiting Reactant : It is the reactant that is completely consumed in the reaction and

which determines the amount of product formed in the chemical reaction.

We are given the mass of both the reactant AgNO₃ and NaOH.

We have to find out the Limiting reactant. Once we figure out the Limiting reactant, we can find the mass of product Ag₂O produced.

Step 1 : Convert 0.200 g of AgNO₃ and 0.200 g of NaOH to moles .

Molar mass of AgNO₃ = 1 * atomic mass of Ag + 1 * atomic mass of N + 3 * atomic mass of O.

= 1 * 107.87g / mol + 1 * 14.007 g/mol + 3 * 15.999 g/mol

= 107.87g/mol + 14.007 g/mol + 47.997 g/mol

= 169.874 g/mol

moles of AgNO₃ = 0.200 g AgNO₃ * 1 mol AgNO₃ / 169.874 g AgNO₃ = 0.001177 mol AgNO₃ .

Molar mass of NaOH = 1 * atomic mass of Na + 1 * atomic mass of O + 1 * atomic mass of H

= 1 * 22.99 g/mol + 1 * 15.999 g/mol + 1 * 1.00794 g/mol

= 22.99 g/mol + 15.999 g/mol + 1.00794 g/mol

= 39.9969 g/mol  

moles of NaOH = 0.200 g NaOH * 1 mol NaOH / 39.9969 g NaOH = 0.005000 mol NaOH .

Step 2 : Calculate the amount of Ag₂O formed by each AgNO₃ and NaOH, considering that the other reactant is in excess.

2 AgNO₃(aq) + 2 NaOH(aq) → Ag₂O(s) + 2 NaNO₃(aq) + H₂O(l)

0.001177 mol AgNO₃ will give = 0.001177 mol AgNO₃ * 1 mol Ag₂O / 2 mol AgNO₃

= 0.0005885 mol Ag₂O

0.005000 mol NaOH will give = 0.005000 mol NaOH . * 1 mol Ag₂O / 2 mol NaOH

= 0.0025 mol Ag₂O

We can see that among AgNO₃ and NaOH, AgNO₃ is the one which gives less moles of

Ag₂O. So AgNO₃ is the limiting reactant and the moles of Ag₂O formed actually is the one we have calculated from AgNO₃ .

Step 3 : Convert 0.0005885 mol Ag₂O to grams.

Molar mass of Ag₂O = 2 * atomic mass of Ag + 1 * atomic mass of O.

= 2 * 107.87 g/mol + 1 * 15.999 g/mol

= 215.74 g/mol + 15.999 g/mol

= 231.739 g/mol

grams of Ag₂O = 0.0005885 mol Ag₂O * 231.739 g Ag₂O / 1 mol Ag₂O

= 0.1364 grams of Ag₂O

In the question, we are given the mass of reactants in 3 significant figure, so our

final answer must also be in 3 significant figure.

Hence the mass of Ag₂O produced is 0.136 grams.