Let’s see how you did! If you didn’t get a few of these, don’t let it stress you out – it just means you need to play with more experiments in this area. We’re all works in progress, and we have our entire lifetime to puzzle together the mysteries of the universe!
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1. Determine the number of moles of N2O4 needed to react completely with 2.56 mol of N2H4 for the reaction
2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(l)
Find the relation between moles of N2H4 and N2O4 by using the coefficients of the balanced equation:
2 mol N2H4 is proportional to 1 mol N2O4
Therefore, the conversion factor is 1 mol N2O4/2 mol N2H4:
moles N2O4 = 2.56 mol N2H4 x 1 mol N2O4/2 mol N2H4
moles N2O4 = 1.28 mol N2O4 (answer)
2. Determine the number of moles of N2 produced for the reaction
2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(l) when the reaction begins with 4.52 moles of N2H4.
Find the relation between moles of N2H4 and N2 by using the coefficients of the balanced equation:
2 mol N2H4 is proportional to 3 mol N2 In this case, we want to go from moles of N2H4 to moles of N2, so the conversion factor is 3 mol N2/2 mol N2H4:
moles N2 = 4.52 mol N2H4 x 3 mol N2/2 mol N2H4
moles N2 = 6.78 mol N2O4 (answer)
3. The balanced equation for the synthesis of ammonia is 3 H2(g) + N2(g) ? 2 NH3(g). From the balanced equation, it is known that:
1 mol N2 ? 2 mol NH3
Use the periodic table to look of the atomic weights of the elements to calculate the weights of the reactants and products:
1 mol of N2 = 2(14.0 g) = 28.0 g
1 mol of NH3 is 14.0 g + 3(1.0 g) = 17.0 g
These relations can be combined to give the conversion factors needed to calculate the mass in grams of NH3 formed from 72.0 g of N2:
mass NH3 = 72.0 g N2 x 1 mol N2/28.0 g NH2 x 2 mol NH3/1mol NH3 x 17.0 g NH3/1 mol NH3
mass NH3 = 87.43 g NH3 (answer)
To obtain the answer to the second part of the problem, the same conversions are used, in a series of three steps:
(1) grams NH3 ? moles NH3 (1 mol NH3 = 17.0 g NH3)
(2) moles NH3 ? moles N2 (1 mol N2 ? 2 mol NH3)
(3) moles N2 ? grams N2 (1 mol N2 = 28.0 g N2)
mass N2 = 3.00 x 103 g NH3 x 1 mol NH3/17.0 g NH3 x 1 mol N2/2 mol NH3 x 28.0 g N2/1 mol N2
mass N2 = 2.47 kg N2 (answer)
4. The most dangerous chemicals in your set are:
a. C1000 & C3000: Potassium Hexa-cyanoferrate(II) – do not release this back into the environment, as it is harmful to aquatic organisms, , so dispose of in container as directed. Do not inhale the dust, and avoid contact with skin and eyes.
b. C1000 & C3000: Hexamethyl-eneteramine – flammable, do not inhale the dust and avoid contact with skin, always wear protective gloves when handling.
c. C1000 & C3000: Copper Sulfate – wear protective gloves and glasses when handling, very poisonous to aquatic organisms, so dispose of in container as directed. Do not release into environment.
d. C3000: Calcium Hydroxide – do not inhale dust, wear protective gloves and glasses when handling, caustic.
e. C3000: Potassium Permanganate – flammable, wear protective gloves and glasses when handling, very poisonous to aquatic organisms, so dispose of in container as directed. Do not release into environment.
f. C3000: Sodium Hydrogen Sulfate – caustic, wear protective gloves and glasses when handling.
g. C3000: Hydrochloric Acid –wear protective gloves and glasses when handling. You’ll be generating a small amount of a weak solution of this with your experiments, so follow all directions carefully.
h. C3000: Sodium Hydroxide – caustic, wear protective gloves and glasses when handling. You’ll be generating this with your experiments, so follow all directions carefully.
5. After mixing two chemicals together, you observe your solution bubbles (generates a gas), gets warm (exothermic) and turns litmus paper red (acidic).
6. If you cut an apple in half and leave it for ten minutes, it turns brown because the fruit is basically rusting. An enzyme in the fruit (polyphenol oxidase) reacts with the oxygen in the air. You can add lemon juice or other acid to slow this chemical reaction down or by removing the oxygen (by vacuum sealing the fruit). If you cut the apple with a rusty knife, the reaction will occur even faster!
7. The balanced equations are below:
a. 3 KOH + H3PO4 –> K3PO4 + 3 H2O
b. 4 NH3 + 5 O2 –> 4 NO + 6 H2O
c. 2 BF3 + 3 Li2SO4 –> B2(SO3)3 + 6 LiF
8. Let’s figure out how many moles are in 26 grams of CO2. First, we peek at the periodic table and find out the atomic mass of carbon is 12, and the atomic mass for oxygen is 16. Here’s how you find the mass of CO2:
C + 2 (O) –> 12 + 2(16) = 44
So one mole of CO2 weighs 44 grams. This now becomes our conversion factor of (1 mole)/(44 grams) and we use it like this:
Number of moles of CO2 = 26g x (1 mole/44grams) = 0.59 moles
So there are 0.59 moles of CO2 in 26grams.
9. If we have 42 moles of H2SO4, how many grams is that?
First, look up H, S, and O in the periodic table to find their atomic masses: H = 1, S = 32, O = 16. So the atomic mass of H2SO4 is:
H2SO4 –> 2H + S + 4O –> 2(1) + 32 + 4(16) = 98
So one mole of H2SO4 weighs 98 grams. Now use this conversion to find the mass for 42 moles:
Grams of H2SO4 = 42 moles x (98 grams/1mole) = 4.12 kg
So there are 4.12 kg of H2SO4 in 42 moles. (That’s a lot – you’d never use that much!)
10. Redox reactions involve electron transferring between atoms. Hydrogen ions (protons) transfers are found in acid-base reactions.
11. Grind up the reactants into a fine powder, swirl them in a minimal amount of water, raise the temperature of the water, and add an appropriate catalyst.
Here’s why: Increasing the temperature will usually increase the rate of reaction, as the higher the temperature of the reactants, the more kinetic energy is in the system. Putting the reactants in a solution allows them to not only react on all sides of the substance but also allows the reactants the most amount of freedom to mingle and react with each other. A higher concentration of the reactants increases the reaction rates because the number of collisions increase. Fine powders react more quickly than large chunks because there’s more area exposed to react when a substance is in powder form, so surface area plays a role in the reaction rates as well. To speed up a reaction without altering the chemistry of the reaction involves adding a catalyst.
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