Hydrogen sulfide reacts with oxygen to give sulfur dioxide...

Hydrogen sulfide reacts with oxygen to give sulfur dioxide and water according to the following balanced equation: 2 H2S (g) + 3 O2 (g) ? 2 SO2 (g) + 2 H2O (g) If we allow 8.52 g of hydrogen sulfide (H2S) to react with 9.15 g of oxygen gas (O2): a) Calculate the number of moles of SO2 that can be produced from this reaction (this is a limiting reactant problem): b) Calculate the mass in grams of SO2 that can be produced from this reaction: c) Which is the limiting reactant in this reaction? Show your work, use correct units and significant figures in your results.

Transcript

00:01 Hello students in this question we have given an equation where 2h2s plus 3o2 reacts to form 2 s o2 plus 2 h2o and we have given that is 8 .5 4 gram of h2 is taken and 9 .15 gram of o2 is taken and in part a we we have asked about the number of the moles of so2 produced.

00:40 So to answer this, first we will know about the limiting reagent in the reaction.

00:48 Limiting reagent in the reaction.

00:53 To check the limiting reagent, first we need to rewrite the equation and find out the amount of the reactant reacted to produce so2.

01:06 So first we will rewrite the reaction and just checking about the amount of the like 2 h2 s is 34 grams smaller mass of h2s 3 multiplied by molar mass of the o2 that is 32.

01:29 It's come out to be 68 and it's come out to be so what we have saw that here we can say 68 gram of h2s react with with 96 gram of o2 or what we can say just oppose it or 96 gram of o2 or 96 gram of o2 needs 68 gram of h2s and what we have given the amount of the o2 is 1 .95 gram of o2 is given so it requires 68 divided by 96 multiplied by 9 .15 which comes out to be 6 .48 grams of h2s used therefore therefore, what we can say that from this, we can conclude that o2 is a h2s will be left.

02:57 Given amount minus used amount.

03:01 Left amount is left amount of h2s is equal to given amount is 8 .52 and used amount is 6 .48.

03:16 So from this we can conclude that o2 is a limiting reagent.

03:27 So reaction will be proceed according to o2.

03:33 Now we need to find the about moles of so2 produced.

03:44 To check about the moles of the so2 produced, we have just we need to calculate that 2h2 we write the equation 2h2s plus 302 gives out 2 so2 plus 2h2.

04:05 Here we have saw that 68 gram of h2os will produce 128 gram of so2.

04:18 What we can say, 68 gram of h2s produced 1 .208 gram of h2 produced 1 .2.

04:31 128 grams of so2 therefore 6 .48 gram of h2 will produce that is equal to 128 divided by 68 multiplied by 6 .48.

04:52 So it's come out to be 12 .19 grams of so2...

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