Determine the balanced equation for the neutralization equatio...

Question

Determine the balanced equation for the neutralization equation

Ca(OH)₂ (aq) + HCN (aq) →

Accepted Answer

Here in this practice question, it says determine the balanced equation for the neutralization equation. Now before we actually do it, realize that this is a neutralization equation because it possesses an acid and a base. The acid is in the form of hydrocyanic acid and the base is calcium hydroxide. Alright. Now step 1, we have to break it up into its ionic forms, both ionic compounds. So calcium hydroxide possesses calcium which is in group 2a, so its charge is 2 +. Hydroxide is a polyatomic ion and its charge is minus 1. Here we have our acid. Now the acid will break up into h+ 1, and then c n minus, which is your cyanide ion. Now remember at this point, opposites attract. Opposite charges will attract one another. When the numbers in the charges are different though, they don't just cancel out, they crisscross. So 2 comes here and then one comes here. So that's gonna give me c a. Remember, we write the positive ion first, and then the 2 is for the entire c n ion, so put it in parentheses and then a little 2. Alright. Plus, then we have here, we have plus 1 and minus 1, the numbers are the same, so when they combine they just cancel out. Now, h+ and o h minus, think about it. When they combine, it gives us h o h which can be written as water. And remember, water in these molecular equations will exist as a liquid. So just remember, if you form h o h, that's just water. So we know reaction has definitely occurred because if you create a solid, a gas or liquid water, we know a reaction has occurred. Alright. So we know water is a liquid, but what about calcium cyanide? Here, calcium doesn't exactly have its own solubility rule. It's just part of exceptions. Cyanide ion isn't even talked about in terms of solubility rules. And because it's not stated within solubility rules, we can assume that it's going to be a solid. Now we have our equation but we have to still balance it. If we look, we have 1 calcium, 1 calcium. We have, 2 hydroxides here but we don't see hydroxide exactly within this, product side. That's because it's part of water. So let's just not worry about hydroxide for right now. For c n, we have 1 c n here but 2 over here. Right? 2 of them over here. So let's put a 2 here. Alright. Now let's go back. If we look we have 2 hydroxides, right, being distributed here so that means we have 2 oxygens there and we have 2 hydrogens. This 2 is getting distributed to this h, so that also means we have another 2 hydrogens. So that's 4 hydrogens total on the product side. So if we go to the, on the reactant side, if we go to the product side, we would have to place a 2 right here because that 2 would get distributed to the h's to give us 4 hydrogens, and to the oxygens to give us 2 oxygens. And we would see that both sides would match. So here this would be 1212 for the coefficients of our balanced molecular equation. Now if you want more in-depth steps in terms of balancing chemical equations, make sure you go back and take a look at my topic videos on balancing chemical equations. At this point, we've done enough of these that I feel like we've mastered it, so so we're not going through every single step with it. And also, these questions are pretty basic or simple in terms of balancing these chemical equations, but if you're still having trouble, make sure you go back and take a look at those earlier videos to keep on practicing. So at this point, we know how to balance molecular equations. Just remember to rely on the rules to guide you to your answer.

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Determine the balanced equation for the neutralization equation
Ca(OH)₂ (aq) + HCN (aq) →