Answer: acetone molecule ( CH₃-CO-CH₃)
Explanation:
1) Acetone is CH₃-CO-CH₃
2) That is a molecule (build up of covalent bonds).
3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.
This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.
That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).
The molar mass of CO2 is 44 grams per mole.
165 grams / 44 grams per mole of CO2 = 3.75 moles CO2
Using Avogadro’s law where 1 mole of substance equals
6.023 x 10^23 molecules
3.75 moles CO2 (6.023 x 10^23 molecules /mole) = 2.26 x 10^24 molecules CO2
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
![2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\](https://tex.z-dn.net/?f=2Al%28s%29%2B3S_2%28g%29%5Crightarrow%202Al_2S_3%28s%29%5C%5C)
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
![n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2](https://tex.z-dn.net/?f=n_S%5E%7Bavailable%7D%3D1.00gS_2%2A%5Cfrac%7B1molS_2%7D%7B64gS_2%7D%20%3D0.0156molS_2%5C%5Cn_S%5E%7Bconsumed%5C%20by%20%5C%20Al%7D%3D1.00gAl%2A%5Cfrac%7B1molAl%7D%7B27gAl%7D%2A%5Cfrac%7B3molS_2%7D%7B2molAl%7D%3D0.0556molS_2)
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
![m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3](https://tex.z-dn.net/?f=m_%7BAl_2S_3%7D%3D0.0156molS_2%2A%5Cfrac%7B2molAl_2S_3%7D%7B3molS_2%7D%20%2A%5Cfrac%7B150gAl_2S_3%7D%7B1molAl_2S_3%7D%20%3D1.56gAl_2S_3)
c) The leftover is computed as follows:
![m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\](https://tex.z-dn.net/?f=m_%7BAl%7D%5E%7Bexcess%7D%3D%280.0556-0.0156%29molS_2%2A%5Cfrac%7B2molAl%7D%7B3molS_2%7D%2A%5Cfrac%7B27gAl%7D%7B1molAl%7D%20%3D0.72%20gAl%5C%5C)
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.