The bold with red X\'s indicate the answer. Can someone explainhow to get the fi
ID: 691062 • Letter: T
Question
The bold with red X's indicate the answer. Can someone explainhow to get the first part?
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Thanks. Here^'s the question: If iron is oxidized to Fe^2+ by a copper(II) sulfatesolution, and 0.340 grams of iron and11.5 mL of 0.526M copper(II)sulfate react to form as much product as possible, how manymillimoles (mmol) of the non-limiting reactant will remain unusedat the end of the reaction? Amount of non-limiting reactant remaining unused = 3.87E-2 mmol Copper(II) sulfate in solution is blue in color. Iron(II) sulfateis colorless. In the reaction described above, will the solution atthe end of the experiment be blue or colorless? The solution will be colorless The bold with red X^'s indicate the answer. Can someone explainhow to get the first part? I^'ll be sure to rate! Thanks.
Explanation / Answer
Fe + CuSO4 ----> FeSO4 + Cu No . of moles of CuSO4 , n = Molarity * Volume =0.526 M * 0.0115 mL =0.006049 moles atomic mass of Fe is = 55.8 g 55.8 g of Fe reacts with 1 moles of CuSO4 X g of Fe reacts with 0.006049 moles ofCuSO4 X = 0.006049* 55.8 = 0.3375 g So 0.340 - 0.3375 = 0.0025 g of Fe left unreacted which is theexcess reactant & CuSO4 is the limiting reactant No . of moles of non-limiting reactant , n = mass / atomicmass of Fe = 0.0025 g / 55.8 g = 4.48 * 10^-5 moles =4.48 * 10^-2 * 10^-3 moles = 0.0448 millimoles Since all the copper sulphate is consumed so the colour of thefinasl solution is colourless.Related Questions
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