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Chemistry lab - limiting reagent - Can someone verify?
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[QUOTE="John Ker, post: 6066368, member: 636966"] [h2]Homework Statement [/h2] Suppose that, for [I]reaction 4,[/I] you could not find the bottle of 7 M H2SO4 so you added 5.00 mL of the 1.00 M H2SO4 instead. How would this impact your final yield of Copper. (Show with calculations how this would impact the limiting reagent.) [h2]Homework Equations[/h2] [I]CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)[/I] [h2]The Attempt at a Solution[/h2] This was the series of transtions done to the copper. [I]Cu(NO3)2(aq) + 2NaOH(aq) --> Cu(OH)2(s) + 2 NaNO3(aq)[/I] [I]Cu(OH)2(s) + heat [I]--> [/I] CuO(s) + H2O(l) CuO(s) + H2SO4(aq) [I]--> [/I] CuSO4(aq) + H2O(l) CuSO4(aq) + Zn(s)--> ZnSO4(aq) + Cu(s)[/I] I initially started with 10.00ml of .4M Cu(NO3)2. Hence .004 moles. The ratio for each copper reactant to the desired copper product is 1:1. Hence we will end up with .004 CuSO4. Now, in the case of using 7M H2SO4, the CuSO4 is the limiting reagent. However, we use 1M * .005ml = .005mol, doesn't this also keep CuSO4 as the limiting reagent, having no effect on the final yield of copper? Thanks! [/QUOTE]
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Chemistry lab - limiting reagent - Can someone verify?
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