You are given the following reaction equation.
2 Fe(aq) + 3 CuCl2(aq) --> 2 FeCl3(aq) + 3 Cu(s)
If 0.0125 moles of iron solid reacted, how many moles of copper solid would be produced ?
Put your answer in the form of a.bc x 10-d and input your values for a,b,c and d with no spaces or punctuation.
+15000 If 0.0125 moles of iron solid reacted, how many 2 moles of copper solid would be produced ?
Hello Guest!
\(2 Fe + 3 CuCl_2 --> 2 FeCl_3 + 3 Cu\)
relative atomic mass
Fe 55.85
Cu 63.54
\(2\cdot55.85g\ Fe\ \widehat =\ 3\cdot 63.54g\ Cu\\ 111.7g\ Fe\ \widehat =\ 190.62g\ Cu\\ 0.0125\ mol:2\ mol=x:190.62g\ Cu\\ x=\frac{190.62\cdot 0.0125}{2}\ g\ Cu\\ x=1.91375g\ Cu\cdot \frac{mol}{63.54g}\)
\(x=1.875\times 10^{-2}mol\ Cu\)
\({\color{blue}1.875\times 10^{-2} moles\ of\ Copper}\ solid\ would\ be\ produced.\)
!
+2511 Hello Asinus:
Your solution is correct, but your method is overkill. The question introduces a molar ratio and asks for the number of moles. Molar mass is not needed here, only a basic ratio formula.
Question (restated in active voice): Two (2) moles of Fe (iron) react to produce three (3) moles of Cu (copper). How many moles of copper precipitate, if 0.0125 moles of iron react?
Solution:
\(\text {Molar ratio: 2} \small \mathrm {Fe} : \normalsize {3} \small \mathrm{Cu } = \normalsize \text {0.0125} \small \mathrm{Fe} : \normalsize \text {X} \small \mathrm {Cu } \normalsize \text {; solve for X.}\\ 0.0125 \small \mathrm {Fe} * \normalsize \dfrac{3 \small \mathrm {Cu}} {\normalsize{2} \small \mathrm {Fe}} = 0.01875 \small \mathrm {Cu} \normalsize \rightarrow 1.875 \;x\; 10^{-2}\; \mathrm {moles \; Cu} \)
GA
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+15000 Oh wow !
Hello GingerAle!
I made a masterpiece there. Think crookedly and still get the right result! I woke up at 3 a.m. and never fell asleep again. I turned too much to chemistry and not so much to mathematics. Sorry, it was three o'clock in the morning. I recommend the studends your method! Thanks for the simple solution!
!
