Asked by Abby
I have to balance an equation and I had to make the equation from an experiment we did.
So far, my equation looks like this and I know what I have is correct:
Cu(NO3)2+H20+NaOH----->Cu(OH)2
I think I probably need an N or a Na on the right side. What do you think?
You don't need the water (at least you don't need it to balance the equation), Cu(OH)2 is correct. You need NaNO3 on the right. If you want to show the water, without worrying about making everything balance with it in place, make Cu(NO3)2(aq) + NaOH(aq) ==> Cu(OH)2(s) + NaNO3(aq)
That shows the water is present (from the aq part without having to balance it. AND it shows the Cu(OH)2 is a ppt in the aq solution. The equation I wrote above is not balanced yet. I think it needs a 2NaNO3.
Thanks, that all makes sense. Just one thing: How do you know I need NaNO3 on the right?
Cu(NO3)2 + NaOH ==>
This a double replacement reaction (double exchange reaction). If the Cu goes with the OH, then the Na goes with the NO3^-. You simply change the + ions with the - ions.
Oh, duh, thanks!
So far, my equation looks like this and I know what I have is correct:
Cu(NO3)2+H20+NaOH----->Cu(OH)2
I think I probably need an N or a Na on the right side. What do you think?
You don't need the water (at least you don't need it to balance the equation), Cu(OH)2 is correct. You need NaNO3 on the right. If you want to show the water, without worrying about making everything balance with it in place, make Cu(NO3)2(aq) + NaOH(aq) ==> Cu(OH)2(s) + NaNO3(aq)
That shows the water is present (from the aq part without having to balance it. AND it shows the Cu(OH)2 is a ppt in the aq solution. The equation I wrote above is not balanced yet. I think it needs a 2NaNO3.
Thanks, that all makes sense. Just one thing: How do you know I need NaNO3 on the right?
Cu(NO3)2 + NaOH ==>
This a double replacement reaction (double exchange reaction). If the Cu goes with the OH, then the Na goes with the NO3^-. You simply change the + ions with the - ions.
Oh, duh, thanks!
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