1) The formation of water is represented by the following equation :
2H2(g) + O2(g) -> 2H2O(g)
a.What is the limiting reactant if 4 mol of oxygen reacts with 16 mol of hydrogen?
b.What amount (in moles) of water is produced in this reaction?
2) . Sliver nitrate, AgNO3(aq),reacts with iron (III) chloride,FeCl3(aq),to producer silver chloride, AgCl(s), and iron(III) nitrate, Fe(NO3)3(aq).
3AgNO3(aq) + FeCl3(aq) -> 3AgCl(s) + Fe (NO3)3(aq)
a.If solution containing 18.00 g of silver nitrate is mixed with a solution containing 32.4 g of iron(III) chloride, which is the limiting reactant?
b.What amount in moles of iron (III) nitrate is produced in this reaction?
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I answered this type question for your post under Nikki. We prefer you to use the same screen name. It helps us answer your question faster.
well im not nikki lol...we justhappen to ...
Whatever. Just look at the way I solved that type problem for Nikki.
1) a. To determine the limiting reactant, we need to use the mole ratio between the reactants and compare it with the actual amount of each reactant given.
Given:
- 4 moles of oxygen (O2)
- 16 moles of hydrogen (H2)
The balanced equation tells us that the ratio between O2 and H2 is 1:2. So, we calculate the moles of oxygen required to react with the given 16 moles of hydrogen:
16 moles H2 x (1 mole O2 / 2 moles H2) = 8 moles O2
Since we have only 4 moles of oxygen, which is less than the required 8 moles, oxygen is the limiting reactant.
b. We can find the amount of water produced using the mole ratio between the limiting reactant (oxygen) and the product (water) from the balanced equation:
From the balanced equation: 2H2(g) + O2(g) -> 2H2O(g)
The mole ratio between O2 and H2O is 1:2.
Using the 4 moles of oxygen, we can determine the moles of water produced:
4 moles O2 x (2 moles H2O / 1 mole O2) = 8 moles H2O
Therefore, 8 moles of water are produced in this reaction.
2) a. To determine the limiting reactant, we need to compare the amounts of silver nitrate (AgNO3) and iron(III) chloride (FeCl3) based on their molar masses.
Given:
- 18.00 g of silver nitrate (AgNO3)
- 32.4 g of iron(III) chloride (FeCl3)
First, calculate the number of moles for each substance using their molar masses:
- Silver nitrate (AgNO3): molar mass = atomic mass of Ag + atomic mass of N + (3 x atomic mass of O)
= 107.87 g/mol + 14.01 g/mol + (3 x 16.00 g/mol) = 169.87 g/mol
Moles of AgNO3 = mass / molar mass = 18.00 g / 169.87 g/mol ≈ 0.106 moles
- Iron(III) chloride (FeCl3): molar mass = atomic mass of Fe + (3 x atomic mass of Cl)
= 55.85 g/mol + (3 x 35.45 g/mol) = 162.20 g/mol
Moles of FeCl3 = mass / molar mass = 32.4 g / 162.20 g/mol ≈ 0.200 moles
Comparing the moles of each reactant, we find that silver nitrate has fewer moles than iron(III) chloride, indicating that silver nitrate is the limiting reactant.
b. To determine the amount of iron(III) nitrate (Fe(NO3)3) produced, we again refer to the balanced equation:
From the balanced equation: 3AgNO3(aq) + FeCl3(aq) -> 3AgCl(s) + Fe (NO3)3(aq)
The mole ratio between AgNO3 and Fe(NO3)3 is 3:1.
Using the 0.106 moles of silver nitrate (the limiting reactant), we can calculate the moles of iron(III) nitrate produced:
0.106 moles AgNO3 x (1 mole Fe(NO3)3 / 3 moles AgNO3) ≈ 0.0353 moles Fe(NO3)3
Therefore, approximately 0.0353 moles of iron(III) nitrate are produced in this reaction.