You react 25.0 liters of oxygen gas at STP with 50.0 ml of ethanol liquid. Ethanol has a density of 0.789 g/ml.
The excess reactant is:
-C2H5OH
-O2
-CO2
-H2O
-C
-H2
-CO
I do these the long way.
mols O2 = 25/22.4 = estimated 1+.
g ethanol = m = vd = 50.0 x 0.789 = estimated 40 g, then
mol = g/molar mass = estd 40/46 = estd 0.9
C2H5OH + 3O2 ==> 2CO2 + 3H2O
Convert mols O2 to mols CO2 using the coefficients in the balanced equation.
1+mol O2 x (2 mols CO2/3 mol O2) = estd 0.7
Convert mols ethanol to mols CO2.
0.9 x (2 mols CO2/1 mol ethanol) = estd 1.8.
The two values for mols CO2 produced don't agree which means one of them is wrong; in limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent. Therefore, the limiting reagent is O2 which makes the excess reagent .....?
You should confirm all of those numbers because I've estimated them but I don't think it will change the answer.
ELASTICITY, LIKE YOUR SOCKS (10/10 points)
The crystal structure of graphite is shown below. Use the figure to answer the following questions.
Compare the Young's moduli for the directions indicated in the figure. Fill in the blank.
The modulus along a is _________ that along b
equal to - correct
The modulus along a is _________ that along c
greater than - correct
The modulus along b is _________ that along c
greater than - correct
Which of the following reasons best explain your reasoning when comparing the moduli of b and c?
The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical
Fundamental bending and stretching of bonds is different in the two directions
Van der Waals bonding has a lower bond strength than colavent bonding - correct <--
Stretching of covalent bonds requires greater force than bending such bonds
The opportunity to stetch rings in one direction gives one direction a lower modulus
Which of the following reasons best explain your reasoning when comparing the moduli of a and b?
Stretching of covalent bonds requires greater force than bending such bonds
The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical - correct <---
Fundamental bending and stretching of bonds is different in the two directions
The opportunity to stetch rings in one direction gives one direction a lower modulus
Van der Waals bonding has a lower bond strength than colavent bonding
To determine the excess reactant in a chemical reaction, we need to compare the stoichiometric ratio of the reactants with the actual amounts used.
First, let's find the number of moles of each reactant used:
1. Moles of O2 gas:
Using the ideal gas law equation, we can calculate the moles of O2 gas.
PV = nRT,
where P is the pressure (standard pressure is 1 atm), V is the volume in liters (25.0 L), n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature (standard temperature is 273 K).
Using the equation, we have:
n = (PV) / (RT)
n = (1 atm * 25.0 L) / (0.0821 L·atm/mol·K * 273 K)
Calculating this out, we find that n = 1.09 mol of O2 gas.
2. Moles of C2H5OH (Ethanol):
To find the number of moles of ethanol, we need to use its density and volume.
First, we convert the volume of ethanol from milliliters to liters:
50.0 ml = 50.0 ml * (1 L / 1000 ml) = 0.0500 L
Next, we can calculate the mass of ethanol using the density:
mass = density * volume
mass = 0.789 g/ml * 0.0500 L
mass = 0.0395 g of ethanol
Now, we need to convert the mass of ethanol to moles using its molar mass:
Molar mass of C2H5OH = (2 * Molar mass of C) + (6 * Molar mass of H) + (1 * Molar mass of O)
Molar mass of C2H5OH = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
Molar mass of C2H5OH = 46.08 g/mol
moles of C2H5OH = mass / molar mass
moles of C2H5OH = 0.0395 g / 46.08 g/mol
moles of C2H5OH ≈ 0.000857 mol
Now that we have the moles of each reactant, we can compare the stoichiometric ratio using the balanced chemical equation for the reaction. However, you haven't provided the balanced equation, so we cannot determine the excess reactant without it.
Please provide the balanced chemical equation for the reaction so that we can determine the excess reactant.