When a 1.350g sample of element X is heated in a stream of chlorine gas 6.67g of XCl3 are formed.1.) What(write) is a balance equation for the reaction using 'X' for the element? 2.) Calculate the atomic weight of X. 3.)What mass of X is left unreacted? 4.)When 1.00 grams of X react with excess Cl2 4.27 grams of XCL3 are formed.What is the % yield for this experiment?
2X + 3Cl2 ==> 2XCl3
1.350..y......6.67
mass Cl2 combined must be 6.67-1.35 = 5.29g
Then I would convert grams to moles. moles = grams/molar mass.
Cl2 = 5.29/35.44 = 0.1501
X = 1.350/molarmass = ??
If the formula in the problem is given as XCl3, we know then that ??moles x must be 0.15/3 = 0.05
Now, 1.35/molar mass = 0.05 and solve for molar mass. I get 26.97 but I've estimated here and there and you should do it more accurately. For the last part, 1.00 g X = 1.00/26.98 = 0.0371. Convert to moles XCl3 and that will be 0.0371 x (2 moles XCl3/2 moles X) = 0.0371 moles XCl3. Now convert to grams.
g = moles x molar mass = 0.0371 x 133.34 = 4.94 g.This is the theoretical yield.
%yield = (actual yield/theoretical yield)*100 = ??
1.) To write a balanced equation for this reaction, we need to determine the chemical formula of element X. Since chlorine gas (Cl2) reacted with X to form XCl3, we know that X must have a valency of 3.
The balanced equation for the reaction is:
2X + 3Cl2 -> 2XCl3
2.) To calculate the atomic weight of element X, we need to use the information given in the problem:
Mass of X = 1.350g
Mass of XCl3 formed = 6.67g
From the balanced equation, we can see that there is a 2:2 ratio between X and XCl3. This means that for every 2 moles of X, 2 moles of XCl3 are formed.
Using the given masses, we can calculate the number of moles of X and XCl3:
Moles of X = Mass of X / Atomic weight of X
Moles of XCl3 = Mass of XCl3 / Molar mass of XCl3
Since there is a 2:2 ratio between X and XCl3, the moles of X and XCl3 should be equal. Setting up an equation:
(Mass of X / Atomic weight of X) = (Mass of XCl3 / Molar mass of XCl3)
Rearranging the equation to solve for the atomic weight of X:
Atomic weight of X = (Mass of X * Molar mass of XCl3) / Mass of XCl3
Substituting the given values:
Atomic weight of X = (1.350g * Molar mass of XCl3) / 6.67g
Note: The molar mass of XCl3 should be calculated based on the atomic masses of X and Cl.
3.) To determine the mass of X that is left unreacted, we need to calculate the amount of XCl3 that was formed and subtract it from the initial mass of X.
From the previous calculations, we know that 6.67g of XCl3 is formed from 1.350g of X. So, the mass of X that reacted is 1.350g - 6.67g = -5.32g.
It is important to note that the negative mass result indicates that there was an excess of chlorine gas in this reaction, and X is entirely consumed.
4.) To calculate the percentage yield, we need to compare the actual yield (4.27g) with the theoretical yield, which can be calculated from the balanced equation.
From the balanced equation, we know that the stoichiometric ratio between X and XCl3 is 2:2. This means that for every 2 grams of X, we would expect to produce 2 grams of XCl3.
Using this ratio, we can calculate the theoretical yield of XCl3:
Theoretical yield of XCl3 = (2g XCl3 / 2g X) * 1g X
Now we can calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) * 100
Substituting the given values:
Percentage yield = (4.27g / Theoretical yield) * 100