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Hello students.
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So, in the following video, we have six parts.
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In the first part, we have the following reaction where the combustion of ethane takes place in presence of oxygen.
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So here we have to find the mole ratio of ethane and water.
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So, as you can see, the mole ratio is 2 is to 6 or we can write it as 1 is to 3, which is option number c.
00:25
So, for the next part, we have the following reaction and we have 12 .8 grams of aluminum.
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We have to find the amount of hcl produced.
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So, first we will find the number of moles of aluminum that is weight by molar mass that is equals to 0 .47.
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So, from the equation, we can see that 2 moles aluminum is equivalent to 6 moles hcl.
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So, that means 0 .47 mole aluminum will be equivalent to 6 by 2 into 0 .47 mole hcl that is equals to 1 .42 mole.
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To convert this to mass, we will multiply it with molar mass of hcl that is 36 .5 gram.
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So, that is equals to 51 .9 gram, which is option number a.
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Let's move on to the next.
01:17
So, in the next part, we have 2 moles of c3h6 reacting with oxygen and undergoing combustion.
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We have 48 grams of c3h6 and we have to find the amount of co2 produced.
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So, firstly, we will find the number of moles that is 48 weight by molar mass that is 42 that is equals to 1 .14.
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So, 2 moles c3h6 is equivalent to 6 moles co2.
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So, that means 1 .14 moles c3h6 is equivalent to 6 by 2 into 1 .14 moles co2 that is 3 .42 moles that is equals to 3 .42 into molar mass of co2 that is 44 gram that is equals to 150 .5 gram approximately 151 gram which is option number d...