Consider the following reaction:
H2O(l) + 3NO2(g) --> 2HNO3(l) + NO(g)
What volume of NO (in L) is produced starting from 1.80 g of water and 11.5 g of NO2 at 300 K and 1 atm? (use R = 0.0821 L atm/mol K).
I am completely confused :(
This is a limiting reagent problem. How do I know that? Because BOTH reactants are given. The way you solve these is to wrote a stoichiometry problem using each reagent, see how much of the product will be formed, then take the smaller value as the correct number. Here is a worked example of a stoichiometry problem. Work it once (to the mole stage of the product) with the first reagent, then a second time (to the mole stage) using the second reagent. When the numbers don't agree, always use the smaller value.
Here is the example, just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Don't worry, I'll help you through the problem step by step.
To solve this problem, we can use the concept of stoichiometry, which relates the amounts of substances in a chemical reaction.
Let's start by determining the moles of water (H2O) and nitrogen dioxide (NO2) that we have:
Moles of H2O = mass / molar mass of H2O
Molar mass of H2O = 2 * molar mass of H + molar mass of O
The molar mass of hydrogen (H) is approximately 1 g/mol, and the molar mass of oxygen (O) is approximately 16 g/mol.
Moles of H2O = 1.80 g / (2 * 1 g/mol + 16 g/mol)
Now, let's do the calculation to find the moles of H2O.
Moles of H2O = 1.80 g / 18 g/mol = 0.10 mol
Next, we need to find the moles of NO2 using the same approach:
Moles of NO2 = 11.5 g / molar mass of NO2
The molar mass of nitrogen (N) is approximately 14 g/mol, and the molar mass of oxygen (O) is approximately 16 g/mol.
Molar mass of NO2 = 14 g/mol + 2 * 16 g/mol
Now, let's do the calculation to find the moles of NO2.
Moles of NO2 = 11.5 g / (14 g/mol + 2 * 16 g/mol)
Now that we know the moles of H2O and NO2, we can use the balanced chemical equation to determine the ratio between moles of H2O, NO2, and NO.
From the balanced equation:
1 mol of H2O produces 1 mol of NO
3 mol of NO2 produce 1 mol of NO
In the reaction, we have 0.10 mol of H2O and 0.396 mol of NO2 (which is the result of 3 moles of NO2 divided by the balanced stoichiometric coefficient).
From the stoichiometry, we can see that the limiting reagent is H2O because it produces the fewer moles of NO.
Now, we can calculate the moles of NO produced from the moles of H2O.
Moles of NO = 0.10 mol (H2O) * (1 mol NO / 1 mol H2O)
Finally, we can convert the moles of NO into volume using the ideal gas law equation:
PV = nRT
Where:
P is the pressure (1 atm)
V is the volume of the gas (what we are trying to find)
n is the number of moles of the gas (moles of NO)
R is the ideal gas constant (0.0821 L atm / mol K)
T is the temperature (300 K)
Rearranging the equation, we have:
V = (nRT) / P
Substituting the values, we can find the volume of NO:
V = (0.10 mol) * (0.0821 L atm / mol K) * (300 K) / 1 atm
Now, let's calculate the volume of NO.