Question 4-6 refer to : 3Ag(s) + 4HNO3(AQ) = 3AGNO3(aq) + NO(g) + 2h2o(l)
4. How many moles of NO can be produced by the reaction of 54.0g of Ag?
5. How many grams of AgNO3 can be produced by recating completely 88.0g of Ag?
6. State the number of grams of water that can be obtained by the reaction of 324g of Ag with 126g of HNO3
Here is a worked example. Just follow the steps. This will work all of the ones you posted except for 6. That is a limitiing reagent problem and you need to do that slightly differently.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To answer these questions, we need to use stoichiometry, which allows us to relate the amounts of substances in a chemical reaction based on their balanced equation.
First, let's analyze the balanced equation:
3Ag(s) + 4HNO3(aq) → 3AgNO3(aq) + NO(g) + 2H2O(l)
1 mole of Ag reacts with 1 mole of NO, so the mole ratio between Ag and NO is 1:1.
1 mole of Ag reacts with 3 moles of AgNO3, so the mole ratio between Ag and AgNO3 is 1:3.
1 mole of Ag reacts with 2 moles of H2O, so the mole ratio between Ag and H2O is 1:2.
Now let's solve the specific questions:
4. How many moles of NO can be produced by the reaction of 54.0g of Ag?
Step 1: Convert grams of Ag to moles of Ag.
Molar mass of Ag = 107.87 g/mol
Number of moles of Ag = (mass of Ag) / (molar mass of Ag)
Number of moles of Ag = 54.0g / 107.87 g/mol ≈ 0.5005 mol
Since the mole ratio between Ag and NO is 1:1, we can conclude that 0.5005 moles of NO can be produced.
5. How many grams of AgNO3 can be produced by reacting completely 88.0g of Ag?
Step 1: Convert grams of Ag to moles of Ag.
Number of moles of Ag = 88.0g / 107.87 g/mol ≈ 0.8150 mol
Step 2: Use the mole ratio between Ag and AgNO3 (1:3) to calculate moles of AgNO3.
Number of moles of AgNO3 = (number of moles of Ag) × (mole ratio of AgNO3/Ag)
Number of moles of AgNO3 = 0.8150 mol × (3 mol/1 mol) = 2.445 mol
Step 3: Convert moles of AgNO3 to grams of AgNO3.
Molar mass of AgNO3 = 169.87 g/mol
Mass of AgNO3 = (number of moles of AgNO3) × (molar mass of AgNO3)
Mass of AgNO3 = 2.445 mol × 169.87 g/mol ≈ 415.1 g
Therefore, approximately 415.1 grams of AgNO3 can be produced.
6. State the number of grams of water that can be obtained by the reaction of 324g of Ag with 126g of HNO3.
Step 1: Convert grams of Ag to moles of Ag.
Number of moles of Ag = 324g / 107.87 g/mol ≈ 3.00 mol
Step 2: Convert grams of HNO3 to moles of HNO3.
Molar mass of HNO3 = 63.01 g/mol
Number of moles of HNO3 = 126g / 63.01 g/mol ≈ 2.00 mol
Since the mole ratio between Ag and H2O is 1:2, we can conclude that:
Number of moles of H2O = (number of moles of Ag) × (mole ratio of H2O/Ag) = 3.00 mol × 2 mol/1 mol = 6.00 mol
Step 3: Convert moles of H2O to grams of H2O.
Molar mass of H2O = 18.02 g/mol
Mass of H2O = (number of moles of H2O) × (molar mass of H2O)
Mass of H2O = 6.00 mol × 18.02 g/mol ≈ 108.1 g
Therefore, approximately 108.1 grams of water can be obtained.