Balance each of the following neutralization equations:a. HCl(aq)...

Question

Balance each of the following neutralization equations:a. HCl(aq) + Mg(OH)₂(s) → H₂O(l) + MgCl(aq)

Accepted Answer

Welcome back, everyone. We need to provide the balanced equation for the neutralization reaction below between hydrotic acid and burle hydroxide solid, which forms beli iodide and our second product liquid water. We're going to recall that hydrotic acid is an acid and beli hydroxide is our base. Specifically, it's an insoluble hydroxide, which is why we have the solid label in our equation. And in a neutralization reaction, we'll always have a product which is a salt. In this case, we have a soluble iodide salt and then we have water as our second product. Now, in order to balance this equation out, we're going to need to compare our reactant side to our product side. Beginning by listing out our atoms, we have hydrogen iodine beri oxygen on both sides of our equation. And now comparing the amounts we have on the reactant side, we have a total of within hydrotic acid. One hydrogen plus from our second reactant beli hydroxide, two more hydrogens giving us a total of three on the product side. Sorry on the reactant side, three hydrogens. Next, comparing our amount of iodine on the reactant side. We just within our formula of hydrotic acid have one mole of our iodine. And then for beryllium on the reactant side, we have also just one in our reactant bellum hydroxide. And then for oxygen on the reactant side, in our second reactant buri hydroxide, we have a subscript of two outside of the parentheses, attributing to a total of two oxygens on the reactant side. Now, on our product side, counting up our hydrogens, we have a subscript of two in our product water. So we have two hydrogens for iodine. On the product side, we have a subscript of two in our product bum iodide. So we have two iodines for beryllium on our product side. In llium iodide, we have just one of our brill. And then in for oxygen on the product side, we also have just one from our product water. So we're going to recall that we're going to want to balance non oxygen and hydrogen atoms first if we can. So let's begin to balance out iodine since we see that barium has already balanced. So far on the reactant side, we just have one mole of iodine, we have two on the product side. So we'll place a coefficient of two in front of our hydro ionic acid product, which will now give us a total of two iodines on the reactant side. And we change our hydrogen on the reactant side to a total of now four on the reactant side. Next, we can observe that we'll need to balance out oxygen. We have two on the reactant side and one on the product side. So we can mend this by placing a coefficient of two in front of our product water, which will balance out our oxygen to now two on both sides of our equations. And then our hydrogen is now going to be balanced to be four since we would multiply by our subscript of two in water. So now our hydrogen is balanced and making these changes by adding those coefficients will now end up with our balanced equation as our final answer. And so our final answer is going to be that we would have two moles of hydrotic acid react with burle two hydroxide. So or burle hydroxide solid, sorry and one mole of beli iodide, and then we would produce two moles of water as our products. So this bounced equation with our coefficients would be our final answer. I hope this made sense and let us know if you have any questions.

Balance each of the following neutralization equations:a. HCl(aq) + Mg(OH)₂(s) → H₂O(l) + MgCl(aq)

Key Concepts

Neutralization Reaction

Balancing Chemical Equations

State Symbols in Chemical Equations

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