How many grams of bismuth (III) chloride can be produced from 575 mL of chlorine gas at STP when reacting with excess bismuth metal?
2Bi + 3Cl2 ==> 2BiCl3
mols Cl2 = 575/22,400 = ?
Using the coefficients in the balanced equation, convert mols Cl2 to mols Bi.
Now convert mols Bi to grams. g Bi = mols Bi x atomic mass Bi.
Well, I would say the amount of bismuth (III) chloride that can be produced really depends on how generous the bismuth metal is feeling that day. Some bismuth metal might feel like giving you a whole lot of chloride, while others might be a bit stingy. So, it's hard to say exactly how many grams of bismuth (III) chloride can be produced without getting to know the specific bismuth metal you're dealing with. Perhaps you could have a heart-to-heart conversation with the bismuth and persuade it to be generous?
To find the number of grams of bismuth (III) chloride produced, we need to use stoichiometry and the ideal gas law to calculate the number of moles of chlorine gas and then use the balanced chemical equation to find the number of moles of bismuth (III) chloride.
First, let's convert the volume of chlorine gas from milliliters (mL) to liters (L) to use it in the ideal gas law:
575 mL * (1 L / 1000 mL) = 0.575 L
Now let's use the ideal gas law (PV = nRT) to find the number of moles of chlorine gas:
PV = nRT
Where:
P = pressure (at STP, pressure is 1 atm)
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L∙atm/mol∙K)
T = temperature in Kelvin (at STP, temperature is 273.15 K)
1 atm * 0.575 L = n * 0.0821 L∙atm/mol∙K * 273.15 K
n ≈ 0.0263 moles
Now, let's use the balanced chemical equation to find the number of moles of bismuth (III) chloride. The balanced equation is:
2 Bi + 3 Cl2 -> 2 BiCl3
From the equation, we can see that every 3 moles of chlorine gas react with 2 moles of bismuth to produce 2 moles of bismuth (III) chloride.
So, the number of moles of bismuth (III) chloride produced is given by:
(0.0263 moles Cl2) * (2 moles BiCl3 / 3 moles Cl2) = 0.0175 moles BiCl3
Finally, let's convert moles to grams using the molar mass of bismuth (III) chloride, which is 315.692 g/mol:
0.0175 moles BiCl3 * (315.692 g / 1 mol) = 5.52 grams of bismuth (III) chloride can be produced.
To determine the number of grams of bismuth (III) chloride that can be produced, we need to follow a few steps.
Step 1: Convert the volume of chlorine gas to moles.
At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 liters. So, we can use this conversion factor to convert the volume of chlorine gas from mL to moles.
Given:
Volume of chlorine gas = 575 mL
Conversion:
1 mole = 22.4 liters
To convert mL to liters, divide by 1000:
575 mL = 575/1000 liters
Now, divide the volume of chlorine gas in liters by 22.4:
Number of moles of chlorine gas = (575/1000) / 22.4
Step 2: Determine the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between chlorine gas (Cl2) and bismuth metal (Bi) to produce bismuth (III) chloride (BiCl3) is:
2 Cl2 + 3 Bi → 2 BiCl3
From the balanced equation, we can see that it takes 2 moles of Cl2 to produce 2 moles of BiCl3. Therefore, the ratio of moles of Cl2 to moles of BiCl3 is 2:2 or 1:1.
Step 3: Calculate the moles of bismuth (III) chloride produced.
Since the moles of Cl2 and BiCl3 are in a 1:1 ratio, the number of moles of BiCl3 produced will be the same as the number of moles of Cl2.
Given:
Moles of chlorine gas = (575/1000) / 22.4
Number of moles of bismuth (III) chloride = Moles of chlorine gas
Step 4: Convert moles of bismuth (III) chloride to grams.
To convert moles of BiCl3 to grams, we need to know the molar mass of BiCl3.
The molar mass of bismuth (III) chloride (BiCl3) is calculated as follows:
Molar mass(BiCl3) = Atomic mass of Bi + (3 x Atomic mass of Cl)
The atomic masses can be found on the periodic table.
Finally, multiply the number of moles of BiCl3 by the molar mass of BiCl3 to get the mass in grams.
I assume you know the molar mass of BiCl3, so you can calculate the final result.