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Hello student.
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The question is ammonia reacts with oxygen to form nitric oxide and water vapor.
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So the reaction is 4 moles of ammonia react with 5 moles of o2 that can form 4 moles of nitric oxide and 6 mol of water.
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So when 20 gram of ammonia and 50 gram of o2 are allowed to react which is the limiting region or reactant in this reaction.
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So we have given the mass of nh3 that is 20 gram and the mass of oxygen that is 50 gram so first of all we can calculate the moles of oxygen and ammonia so the formula we can use number of moles so the number of moles is equal to given mass divided by the molar mass.
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So the moles of ammonia that is mass over molar mass.
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So we get this value 1 .17 and similarly we can calculate the moles of oxygen that is mass over molar mass.
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So we get this value is 1 .56.
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So 1 .56 moles of oxygen and 1 .17 moles of ammonia.
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Okay.
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So from the stochometry, mold ratio, we can say that this 4 moles of ammonia react with 5 moles of oxygen that can form 4 mol of nitric oxide and 6 mol of water.
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So now from equation we can say that this 4 mole of ammonia react with the 5 moles of o2 okay so now the moles of ammonia that is 1 .17 that we calculated already so that can react with how many moles of oxygen so 5 by 4 into 1 .17 moles of o2 so we get this value 1 .46 moles of o2 so that mean this 1 .46 moles of o2 consume in this reaction and the moles of o2 present that is 1 .56 and moles of o2 consume in this reaction that is 1 .46 so the moles left that is 1 .56 minus moles of o2, so moles of o2 left 1 .56 minus 1 .46...