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1

Limiting reagent is the reactant used which produces lesser amount of the product as compared to other reactants if other reactants are used in sufficient amounts. Thus it limits the quantity of the product. Now, according to your question, Moles of CH4 = (Mass of CH4 taken)÷(Molecular mass of CH4) = 8/[12+(4×1)] = 8/16 = 0.5 moles of CH4 are ...


1

For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy: The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long. Among the typical competitions are runs about $50$, and $100\,\pu{m}$...


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First, $40$ g of $Al$ is $1.482$ mol $Al$. And $25$ g $Cl_2$ is $0.352$ mol $Cl_2$. You should see that there is too much aluminum. $0.35$ mol $Cl_2$ reacts with $(2/3)·0.352$ mol $Al$ = $0.2347$ mol $Al$. This is less than $1.428$ mol. As you want to obtain the mass of $AlCl_3$, you must see that the number of moles of $AlCl_3$ is the same as the amount ...


2

Sulfur is made of $S_8$ molecules when cooled down from gaseous or liquid state. But when it is obtained by a chemical reaction at room temperature, it is a mixture of many allotropes, like $S_n$, where n is not well defined and may vary. Also, by cooling abruptly liquid sulfur in carbon disulfide and extracting the soluble part, the dissolved sulfur is $...


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Generally, other than the seven diatomic gases at STP ($\ce{H2}$, $\ce{O2}$, $\ce{N2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$... though $\ce{Br2}$ & $\ce{I2}$ have fairly low vapor pressure at room temperature), for convenience, other elements are considered as if they reacted as individual atoms, though, as you state, that is not actually the ...


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I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time. It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the ...


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First, use freshly bought $\ce{NaHCO3}$ as on warming, it converts into more alkaline $\ce{Na2CO3}$. $$\ce{2 NaHCO3 -> Na2CO3 + H2O (g) + CO2 (g)}$$ See this article: 'Vanishing Baking Soda' at the Scientific American. Next, my experience with generating $\ce{CO2}$ is that you can collect it over water, but depending on the level agitation/bubbling, ...


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