The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 

2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then

(i) Which is the limiting reagent?

<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p>The reaction is shown below.</p><p>2A + 4B? 3C + 4D&nbsp;</p><p>According to the above equation, 2 moles of A requires 4 moles of B. So, the number of moles of B required for 5 moles of A is calculated as, </p><p>Moles of B = 5 mol of A * <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>4</mn> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">A</mi> <mi mathvariant="normal"> </mi> </mrow> </mrow> </mfrac> </math> </span>= 10 mol of B</p><p>So, the required number of moles of B is 10 mal but only 6 moles of B are given in the question. Therefore, B is the limiting reagent.</p><p><strong> (ii) calculate the amount of C formed? </strong></p><p>Ans: Now, the amount of C can be calculated by the limiting reagent, that is, the amount of B.</p><p>According to the equation, 4 moles of B gives 3 moles of C . So, the number of moles of C formed by 6 moles of B as follows, </p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Moles of C = 6 mol of B *<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>3</mn> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">C</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">B</mi> <mi mathvariant="normal"> </mi> </mrow> </mrow> </mfrac> </math> </span> = 4.5 mol of C</p>

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