What is the maximum number of mols of copper (III) sulfide that can be formed when 8.0 mols of copper reacts with 9.0 mols of sulfur?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

The chemical reaction is:

$4Cu+3S_2->2Cu_2S_3$

So from the left part of the equation, once can say:

For every 4 moles of copper, 3 moles of sulfur is needed.
For every 8 moles of copper, $x$ moles of sulfur are needed.
For every $y$ moles of copper, 9 moles of sulfur are needed.

$x=(3*8)/4=6$ moles of sulfur We have 9, which is enough

$y=(4*9)/3=12$ moles of copper We have 8, which is not enough

So finally, 8 moles of sulfur will react. The reaction will become:

$4Cu+3S_2->2Cu_2S_3$

$zCu+8S_2->wCu_2S_3$

$z=4*8/3=32/3$

$w=2*8/3=16/3$

$32/3Cu+8S_2->16/3Cu_2S_3$

So the maximum number moles that can be formed are:

$16/3=5,bar(33)g$

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: $Cu_2S(s) + O_2(g) -> 2Cu(s) + SO_2(g)$. What mass of copper(I) sulfide is required in order to prepare 0.650 kg of copper metal?

You have given the stoichiometric equation that represents reduction of cuprous sulfide: $Cu_2S(s) + O_2(g) rarr 2Cu(s) + SO_2(g)$ Sulfur is oxidized; copper and oxygen are reduced. $"Moles of copper"=(650*g)/(63.55*g*mol^-1)=10.3*mol$....

1 Answers 1 views
Copper will react with sulfur to form a copper sulfide, if 1.500g copper reacts with sulfur to form 1.880g of copper sulfide, what is the empirical formula of the copper sulfide?

The idea here is that you need to use the mass of copper and the mass of the copper sulfide to determine how much sulfur the produced compound contains, then...

1 Answers 1 views
The atomic number of sulfur is 16. Sulfur combines with hydrogen by covalent bonding to form a compound, hydrogen sulfide. Based on the number of valence electrons in a sulfur atom, how do you predict the molecular formula of the compound?

We thus predict that hydrogen sulfide would have a formula of $H_2S$, an analogue of the popular $H_2O$ molecule.

1 Answers 1 views
In an experiment, 2.54 grams of copper completely reacts with sulfur, producing 3.18 grams of copper(I) sulfide. What is the total mass of the sulfur consumed?

Every chemical reaction conserves mass. If you finish with $3.18$ $g$ $Cu_2S$, and $2.54$ $g$ of copper metal were used, the balance, the difference between the end mass and the...

1 Answers 1 views
5.6 mols of sodium reacts with 3.4 mols of chlorine. How many mols of sodium chloride are formed? What is the limiting reactant?

$Na(s) + 1/2Cl_2(g) rarr NaCl(s)$ We assume standard conditions. Thus, there are 5.6 mol sodium metal and 3.4 mols $Cl_2$ gas, that is 6.8 mol chlorine atoms . Thus the...

1 Answers 1 views
How many grams of sodium sulfide are formed if 1.50 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 94.0 % yield?

The first step for any chemical reaction is to write out the chemical equation to determine the molar ratios of the reactants and products. Then we will use the molecular...

1 Answers 1 views
How many mols of $"CO"_2$ were added at constant temperature and volume to a $"1-L"$ container containing $"0.400 mols"$ of $"CO"_2$ and of $"H"_2$, and $"0.200 mols"$ of $"CO"$ and of $"H"_2"O"(g)$ ($K_c = 4$) to cause $"CO"_2$ to drop to $"0.300 mols"$?

I got $"0.900 mols CO"_2$ added. At constant temperature and volume, the mols added only affect the partial pressures of each gas and thus their concentrations. This is just...

1 Answers 1 views
If a reaction of sulfur dioxide with dioxygen gas to form sulfur trioxide starts with $"5 mols SO"_2$ and $"5 mols O"_2$, and $60%$ of the $"SO"_2$ was consumed in the reaction, what is the total mols of gas at equilibrium?

$"8.5 mols of gas"$ Write the reaction: $2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)$ For this reaction, you know that the starting mols are: $n_("SO"_2,i) = "5 mols...

1 Answers 1 views
During the course of the reaction, the vessel is found to contain $"7.00 mols"$ of $C$, $"14.0 mols"$ of $H_2O$, $"3.60 mols"$ of $CO$, and $"8.50 mols"$ of $H_2$. What is the reaction quotient $Q$?

The equilibrium given to you looks like this $"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))$ By definition, the reaction quotient, $Q_c$, tells...

1 Answers 1 views
For the reaction $"B"_2"H"_6(g) + 6"Cl"_2(g) -> "B"_2"Cl"_6(g) + 6"H"_2(g)$, if $"0.252 mols Cl"_2$ reacts with $"0.111 mols B"_2"H"_6$, and $DeltaH_(rxn)^@ = -"1396 kJ"$, what is the heat of reaction?

I'm going to guess it has something to do with which reagent you used as the , and the value you used for your of reaction. The limiting reagent, the...

1 Answers 1 views
X