The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Materials<span> and their </span>properties<span>: </span>compounds like<span> sodium chloride - an interactive educational resource for 11 to 14 year olds. ... Elements are substances (</span>like<span> hydrogen and oxygen) that can't be split into simpler substances. ... For </span>each<span> statement, decide whether it describes a mixture or a </span>compound<span> and check the box.</span>
Answer:
C a chemical reaction formed
Explanation:
if something changes smell and nothing else it is a chemical reaction, every other answer does not make sense
Answer:
87.15%
Explanation:
To find percent yield, we can use this simple equation
![\frac{Actual}{Theoretical} *100](https://tex.z-dn.net/?f=%5Cfrac%7BActual%7D%7BTheoretical%7D%20%2A100)
Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.
They give us these values, so to find the percent yield, just plug the numbers in.
![\frac{30.85}{35.40} *100\\\\ =87.15](https://tex.z-dn.net/?f=%5Cfrac%7B30.85%7D%7B35.40%7D%20%2A100%5C%5C%5C%5C%20%3D87.15)
So, the percent yield is 87.15%
An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.