I missed a few days of school due to a sick child. I'm reviewing for an exam and the following questions are giving me trouble as I wasn't there when they were reviewed in class. If possible, I need the answer and step-by-step process so I can see what what to do and make sure I'm doing it right. Thanks.
1.) If you put hydrogen peroxide on a cut, you’ll notice that it bubbles. The bubbles are diatomic oxygen which is effectively “sterilizing” the wound. If you had 10 g of H2O2, how many grams of oxygen would you form on the cut.
2.) Alka Seltzer is a solid mixture of sodium bicarbonate and citric acid that when placed in water undergoes this reaction…..fizzing.
3NaHCO3(s) + C6H8O7(s) = 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
Who is the limiting reactant if you had 100g each of sodium bicarbonate and citric acid?
Bob Pursley answered your first post. It's below.
Dr. Bob, I am unable to see the answer you have provided? I'm not sure why, but I see "Bob Pursley answered your first post. It's below." The problem is, the is no post below.
You must scroll down and find the original post. Here is a link.
http://www.jiskha.com/display.cgi?id=1415924736
1. 2H2O2 ==> 2H2O + O2
mols H2O2 = grams/molar mass = 10/34 = 0.294/
Using the coefficients in the balanced equation, convert mols H2O2 to mols O2.
That's 0.294 mols H2O2 x (1 mol O2/2 mols H2O2) = 0.294 mols H2O2 x 1/2 = 0.147 mols O2.
Now convert mols O2 to grams. g O2 = mols O2 x molar mass O2. The correct answer
0.147 x 32 = 4.70 g O2.
2. This is the same type but asking slightly different question. Try it and see what you can do. Repost with specific questions about what you don't understand if you require more assistance.
Sure! I can help you with these questions. Let's break them down step by step.
1.) If you put hydrogen peroxide on a cut, you'll notice that it bubbles. The bubbles are diatomic oxygen, which is effectively "sterilizing" the wound. If you had 10 g of H2O2, how many grams of oxygen would you form on the cut?
To solve this problem, we need to use stoichiometry. We are given the mass of hydrogen peroxide (H2O2) and we want to calculate the mass of oxygen produced (O2).
The balanced chemical equation for the reaction is:
2H2O2(aq) -> 2H2O(l) + O2(g)
From the equation, we can see that 2 moles of hydrogen peroxide (H2O2) produce 1 mole of oxygen (O2). We need to convert the given mass of hydrogen peroxide to moles and then use the mole ratio to find the mass of oxygen.
Now, let's calculate the number of moles of H2O2:
Molar mass of H2O2 = (2 * atomic mass of hydrogen) + (2 * atomic mass of oxygen)
= (2 * 1.008 g/mol) + (2 * 16.00 g/mol) ≈ 34.02 g/mol
Moles of H2O2 = Mass of H2O2 / Molar mass of H2O2
= 10 g / 34.02 g/mol ≈ 0.294 mol
Since the mole ratio of H2O2 to O2 is 2:1, we can now calculate the mass of oxygen using the mole ratio:
Mass of O2 = Moles of H2O2 * (Mass of O2 / Moles of H2O2)
= 0.294 mol * (32.00 g/mol / 2 mol) ≈ 4.704 g
Therefore, if you had 10 g of H2O2, you would form approximately 4.704 g of oxygen on the cut.
2.) Alka Seltzer is a solid mixture of sodium bicarbonate and citric acid that, when placed in water, undergoes this reaction: fizzing.
3NaHCO3(s) + C6H8O7(s) -> 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
To determine the limiting reactant, we need to compare the moles of sodium bicarbonate (NaHCO3) and citric acid (C6H8O7) and see which one is consumed completely.
First, calculate the number of moles for each substance:
Molar mass of NaHCO3 = atomic mass of sodium + atomic mass of hydrogen + (atomic mass of carbon + 3 * atomic mass of oxygen)
= 22.99 g/mol + 1.008 g/mol + (12.01 g/mol + 3 * 16.00 g/mol) ≈ 84.01 g/mol
Moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3
= 100 g / 84.01 g/mol ≈ 1.19 mol
Molar mass of C6H8O7 = (6 * atomic mass of carbon) + (8 * atomic mass of hydrogen) + (7 * atomic mass of oxygen)
= (6 * 12.01 g/mol) + (8 * 1.008 g/mol) + (7 * 16.00 g/mol) ≈ 192.13 g/mol
Moles of C6H8O7 = Mass of C6H8O7 / Molar mass of C6H8O7
= 100 g / 192.13 g/mol ≈ 0.52 mol
From the balanced equation, we can see that 3 moles of NaHCO3 react with 1 mole of C6H8O7. Therefore, the mole ratio of NaHCO3 to C6H8O7 is 3:1.
Since we have 1.19 moles of NaHCO3 and 0.52 moles of C6H8O7, we can see that NaHCO3 is in excess and C6H8O7 is the limiting reactant. This means that C6H8O7 will be completely consumed, and any excess NaHCO3 will be left over.
Therefore, the limiting reactant in this reaction, if you had 100 g each of sodium bicarbonate and citric acid, is citric acid (C6H8O7).
I hope this helps! Let me know if you have any further questions.