A 1.000g mixture of As4S6(s) and As4S4(s) was burned in excess O2. The products were As4O6(s) and SO2(g). The solid product had a mass of 0.905g. Find the mass % of As4S4(s) in the original mixture.
These problems are two equations in two unknowns that are solved simultaneously.
Let X = mass As4S4
and Y = mass As4S6
------------------------
equation 1 is X + Y = 1.000
The second equation comes from the 0.905 g of the As4O6 product formed.
The equations are
As4S4 + 7O2 ==> As4O6 + 4SO2
As4S6 + 9O2 ==> As4O6 + 10SO2
To save on typing space I'll lt mm stand for molar mass and as stand for atomic mass.
As4O6 formed from As4S4 is
X(mm As4O6/mm As4S4)
As4O6 formed from As4S6 is
Y(mm As4O6/mm As4S6)
If we add those together we come up with equation 2 which is
X(mm As4O6/mm As4S4) + Y(mm As4O6/mm As4S6) = 0.905 g since all of the As4O6 formed has a mass of 0.905 g.
Solve the two equations, find X (you don't need to do Y unless you're interested), then
%As4S4 = (mass As4S4/mass sample)*100 = (X/1.000)*100 = ?
Post your work if you get stuck.
83% as4s4
To find the mass % of As4S4(s) in the original mixture, we need to determine the masses of As4S6(s) and As4S4(s) that reacted.
1. First, let's find the mass of As4S6(s) that reacted.
Mass of As4O6(s) formed = Mass of the mixture - Mass of solid product
Mass of As4O6(s) = 1.000g - 0.905g = 0.095g
2. From the balanced chemical equation, we know that:
2 mol of As4O6(s) is obtained from 1 mol of As4S6(s)
Therefore, 0.095g of As4O6(s) is formed from (0.095g / (212 g/mol)) = 0.000448 mol of As4S6(s)
3. Next, let's find the number of moles of As4O6(s) that formed.
From the balanced chemical equation:
2 mol of As4O6(s) is obtained from 1 mol of As4S6(s)
Therefore, 0.000448 mol of As4O6(s) corresponds to 0.000448 mol / 2 = 0.000224 mol of As4S6(s).
4. Finally, we can calculate the mass of As4S4(s) that reacted.
From the balanced chemical equation, we know that:
8 mol of As4O6(s) is obtained from 1 mol of As4S4(s)
Therefore, 0.000224 mol of As4S6(s) corresponds to (0.000224 mol / 8) * 256 g/mol = 0.007 moles of As4S4(s)
To find the mass % of As4S4(s), we divide the mass of As4S4(s) by the total mass of the mixture and multiply by 100:
Mass % of As4S4(s) = (0.007g / 1.000g) * 100% = 0.7%
Therefore, the mass % of As4S4(s) in the original mixture is 0.7%.
To find the mass % of As4S4(s) in the original mixture, we need to first determine the amount of As4S4(s) that reacted during the combustion and calculate its mass.
1. Start by finding the moles of As4S6(s) and As4S4(s) in the mixture:
- As4S6: The molar mass of As4S6 is 402.24 g/mol. Divide the mass of As4S6 by its molar mass to get the moles of As4S6.
- As4S4: The molar mass of As4S4 is 352.16 g/mol. Divide the mass of As4S4 by its molar mass to get the moles of As4S4.
2. Now, we need to determine the limiting reactant, i.e., which reactant is completely consumed during the combustion. The reactant that produces a smaller amount of one of the products will be the limiting reactant.
- Since As4S6(s) reacts to form As4O6(s), and As4S4(s) reacts to form SO2(g), we need to calculate the moles of product formed from each reactant. This can be done using the stoichiometry of the balanced chemical equation.
- The balanced equation for the reaction is:
As4S6(s) + 15O2(g) -> 4As4O6(s) + 8SO2(g)
- Using the stoichiometry, determine the moles of As4O6(s) formed from the moles of As4S6(s).
- Similarly, determine the moles of SO2(g) formed from the moles of As4S4(s).
3. Compare the moles of As4O6(s) and SO2(g) formed. Whichever has a smaller number of moles would be the limiting reactant.
4. Once you have determined the limiting reactant, calculate the moles of As4S4(s) that reacted based on its stoichiometry with the product formed.
5. Finally, calculate the mass of As4S4(s) that reacted using its moles and molar mass.
6. To find the mass % of As4S4(s) in the original mixture, divide the mass of As4S4(s) that reacted by the initial mass of the mixture and multiply by 100.
By following these steps, you should be able to find the mass % of As4S4(s) in the original mixture.