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Posted

I just had this problem given to me, we just started it and im kind of lost: How many moles of oxygen are consumbed when 96.7 moles of hydrogen sulfide gas are burned producing sulfur dioxide and wator vapor in the process?

 

I got 145 mols of o2, but im not sure

Posted
I just had this problem given to me, we just started it and im kind of lost: How many moles of oxygen are consumbed when 96.7 moles of hydrogen sulfide gas are burned producing sulfur dioxide and wator vapor in the process?

 

I got 145 mols of o2, but im not sure

Firstly, write out the balanced chemical equation. Next, you would normally calculate the number of moles of your given starting reagent, however, since you already have the moles of hydrogen sulfide, you can skip this step.

 

Once you have everything in moles, you can compare directly using the stoichiometry in your balanced chemical equation. I'm trying to be as clear as possible, but without giving away the answer here, though your answer of 145 moles of O2 is incorrect.

 

typically when it says excess, does that involve limiting reactant??/

Indeed, often if one substance is in excess, another will be the limiting reagent.

Posted

Its not. A limiting reagent problem is when you are given all the reactant quantities. For example, if you were given 32 mol of O2 and 96.7 mol HS. However, it is asking how much oxygen is consumed. In other words, it means how much oxygen is needed to fully complete the reaction. If you write out the equation you get

 

2O2 + H2S -> 2H2O + SO2

 

This says that for every mole of hydrogen sulfide (coefficient=1), 2 moles of O2 are required to complete the reaction (coefficient=2). Thus, if you have 96.7 mol H2S, then you would need 96.7 H2S * (2 mol O2 / 1 mol H2S ) = 193.4 mol O2.

 

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Posted
If you write out the equation you get

 

2O2 + H2S -> 2H2O + SO2

 

This says that for every mole of hydrogen sulfide (coefficient=1), 2 moles of O2 are required to complete the reaction (coefficient=2). Thus, if you have 96.7 mol H2S, then you would need 96.7 H2S * (2 mol O2 / 1 mol H2S ) = 193.4 mol O2.

 

Just so you're aware, this answer is also incorrect, as the equation is not yet properly balanced.

Posted

How many grams of zinc chloride solution are required to completely react with 15.0 grams of lithium? the zinc chloride solution is 8.00% zinc chloride by mass?

 

Can somenoe help me figure out where to begin?

Posted

A good starting place is define the equation. The unbalanced equation is ZnCl2 + Li --> LiCl + Zn

 

Essentially the same problem as before except that the masses initially given need to be in terms of moles, and then moles will have to be converted back to mass in the end and then use the mass percent to find the total mass of the solution.

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