I have two questions:
1. A sample of 10.7 g of CO reacts completely with O2 to form CO2. How many grams of CO2 will be formed?
2. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2 g of ammonia are reacted with 787.3 g of carbon dioxide. Which of the two reactants is a limiting reactant.
*My textbook is useless and this is an online class with no lecture. I am supposed to show my work and use the correct number of significant figures. I've been working on this assignment for hours and have hit a wall! Thanks!
#1 is a simple (simple means it isn't a limiting reagent problem) stoichiometry problem. Here is a worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
#2. #2 is a limiting regent problem. The way I solve these is to solve TWO simple stoichiometry problems as in # 1 above. Solve the first one with 637.2 g NH3 to produce H2O and solve the second time with 787.3 g CO2 to produce water. The moles you form from each "problem" will not be the same; the correct answer in limiting reagent problems is ALWAYS the smaller one and the regent producing that value is the limiting reagent.
Post your work if you get stuck.
Hi Dr. Bob,
Thank you for the help! I took a few hours to think about something else and I was finally able to get the problems answered. I hope I did my calculations correctly.
Thanks again!
You're quite welcome.
I'm sorry to hear that you're facing difficulties with your assignment. Don't worry, I'm here to help you understand and solve these questions step by step.
Let's start with the first question:
1. To determine the grams of CO2 formed, we need to use stoichiometry. Stoichiometry is a quantitative relationship between the reactants and products in a chemical reaction.
2. First, we need to write down the balanced chemical equation for the reaction between CO and O2 to form CO2.
The balanced equation is: 2CO + O2 → 2CO2
3. From the balanced equation, we see that 2 moles of CO react to form 2 moles of CO2. Therefore, the mole ratio between CO and CO2 is 2:2 or 1:1.
4. Next, we need to convert the given mass of CO (10.7 g) to moles. To do this, we need the molar mass of CO, which can be calculated using the atomic masses of carbon (C) and oxygen (O) from the periodic table.
The molar mass of C = 12.01 g/mol and the molar mass of O = 16.00 g/mol.
Therefore, the molar mass of CO = (12.01 g/mol) + (16.00 g/mol) = 28.01 g/mol.
Now, we can calculate the moles of CO using the formula: moles = mass / molar mass.
moles of CO = 10.7 g / 28.01 g/mol = 0.382 mol.
5. Since the mole ratio of CO to CO2 is 1:1, the moles of CO2 formed will also be 0.382 mol.
6. Finally, we can convert the moles of CO2 to grams using the molar mass of CO2.
The molar mass of C = 12.01 g/mol and the molar mass of O = 16.00 g/mol.
Therefore, the molar mass of CO2 = (12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol.
grams of CO2 formed = moles of CO2 x molar mass of CO2 = 0.382 mol x 44.01 g/mol = 16.82 g (rounded to two significant figures).
Now, let's move on to the second question:
1. The same approach of stoichiometry will be used to determine the limiting reactant.
2. We are given the masses of ammonia (NH3) and carbon dioxide (CO2). We need to find out which reactant is present in the limiting amount.
3. Begin by writing the balanced chemical equation for the reaction between ammonia (NH3) and carbon dioxide (CO2) to form urea (NH2)2CO and water (H2O).
The balanced equation is: 2NH3 + CO2 → (NH2)2CO + H2O
4. Calculate the moles of ammonia and carbon dioxide using the given masses and molar masses in the same way as done in the previous question.
The molar mass of NH3 = 14.01 g/mol and the molar mass of CO2 = 44.01 g/mol.
moles of NH3 = 637.2 g / 14.01 g/mol = 45.48 mol
moles of CO2 = 787.3 g / 44.01 g/mol = 17.87 mol
5. The mole ratio between NH3 and CO2 in the balanced equation is 2:1. This means that for every 2 moles of NH3, only 1 mole of CO2 is required.
6. To determine the limiting reactant, compare the mole ratio with the moles calculated in the previous step. The reactant that gives the lowest mole ratio is the limiting reactant.
For ammonia (NH3): 45.48 mol / 2 = 22.74 mol CO2 required.
For carbon dioxide (CO2): 17.87 mol.
7. As we can see, ammonia (NH3) has an excess amount compared to the mole ratio. Therefore, carbon dioxide (CO2) is the limiting reactant in this case.
Remember to round your answers to the appropriate number of significant figures as specified in the question or according to the rules of significant figures provided in your class.
I hope this explanation helps you understand how to solve these types of problems. If you have any further questions, feel free to ask!