Balance each of the following neutralization equations:
a. HNO₃(aq) + Ba(OH)₂(s) → H₂O(l) + Ba(NO₃)₂(aq)

Question

Balance each of the following neutralization equations:

a. HNO₃(aq) + Ba(OH)₂(s) → H₂O(l) + Ba(NO₃)₂(aq)

Accepted Answer

Welcome back, everyone. We need to give the correct balanced equation for the neutralization reaction shown below. We have a reaction between periodic acid H I 04 lead to hydroxide in its solid form to form lead to iodate and liquid water as products. We're going to recognize that periodic acid is indeed our acid reacting with an insoluble salt, which is our base and we would form our product, which is another insoluble salt led to ID date. And then we have our second product being water. So when we have an acid and a base reacting and it forming a salt and water as product, we have a neutralization reaction. Now, we need to balance this out. So we're going to want to compare our reactants to our products. So defining the number or the type of atoms that we have on the reactant side, we have hydrogen, iodine oxygen and lead. And we have the same atoms on our product side, hydrogen iodine oxygen and lead. We're going to recall that when we balance chemical equations, we balance non oxygen and non hydrogen atoms first. So defining the number of the types of atoms we have on our reactant side, first, we can count a total of three hydrogen atoms on our reactant side, we can count a total of one iodine atom on the reactant side. A total of six oxygens on the reactant side. And then a total of one lead atom on the reactant side. Now, on our product side, we can count a total of two of our hydrogens. On the product side, we have a total of two of our iodines on the product side. And then we have a total of nine oxygens on the product side given by in our first product led to I date, we would multiply four times our four within our parentheses of ID eight and then times the two As our subscript outside of our parentheses, which would give us eight. And then this would be added to the one oxygen in our second product water, which again gives us nine. Then lastly for lead, we have just one of our lead atoms on the product side. So we can see that our atom lead is balanced already. So our only other non oxygen and non hydrogen atom to balance would be iodine. We only have one mole of iodine on the reactant side and we need two to match with our product side. So we're going to begin by placing a coefficient of two in front of periodic acid, our our reactant, sorry, that's going to change our iodine atom to two. So that they're bounced with the two on the product side. And this will also change our hydrogen to a total of instead of three, we will now have four on the reactant side. And lastly, we will change our oxygen to a total of now 10 on the reactant side. Since we would have two times our subscript four in periodic acid giving us eight oxygens plus the in our second reactant lead to hydroxide. We have oxygen in parentheses. And outside of our parentheses, we have a subscript of two. So that would add 2 to R eight From periodic acid for oxygen, which would give us 10 oxygens total on the reactant side. So now we can observe that we would need to bounce out hydrogen on the product side so that we would have four. And we can get that by placing a coefficient of two in front of our product water, which will change our hydrogen to now four to bounce out with the four on the reactant side. And we will also change our oxygen to a total of now 10 on the product side to balance out with the 10. On the reactant side, We know we have 10 oxygens because we would have our from lead to ID eight, our subscript of four multiplied by our subscript of two, giving us eight plus our coefficient of two in front of water, which would add two more oxygens giving us 10 total. So now with these changes, we've completely balanced out our chemical equation given. So our final answer is going to correspond to two moles of periodic acid reacting with one mole of lead, two hydroxide to form one mole of lead to iate and two moles of liquid water. This is our final answer as the correct bounced chemical equation for our given neutralization reaction. I hope this made sense and let us know if you have any questions.

Balance each of the following neutralization equations:
a. HNO₃(aq) + Ba(OH)₂(s) → H₂O(l) + Ba(NO₃)₂(aq)

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Balance each of the following neutralization equations:a. HNO3(aq) + Ba(OH)2(s) → H2O(l) + Ba(NO3)2(aq)

Verified video answer for a similar problem:

Key Concepts

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Balance each of the following neutralization equations:
a. HNO₃(aq) + Ba(OH)₂(s) → H₂O(l) + Ba(NO₃)₂(aq)