Complete and balance the equation for each of the following react...

Question

Complete and balance the equation for each of the following reactions:c. HCl (aq) + NaHCO₃(s) →

Accepted Answer

Welcome back, everyone. We need to supply the products for the reaction of hydrochloric acid with potassium bicarbonate and balance the chemical equation. Now, we need to analyze our compounds to understand the types of compounds we have in our reactants. So, beginning with potassium bicarbonate, recognize that this is an ionic compound consisting of our potassium ion which is a metal with a plus one charge as a CAD ion and our bicarbonate anion H C 031 minus which is a nonmetal or poly atomic ion. So, ultimately, we can label this as a metal bicarbonate. Now, next, we have hydrochloric acid as our second reactant. We recognize that H C L consists of our H plus which is an acidic proton and then chlorine as our C L minus anion or chloride anion. Now, as a whole, we have two nonmetals bonded to one another. And so therefore, we have a molecular or covalent compound. And again, H C L is representative as an acid recall that it's one of our memorized strong acids from our textbooks. So whenever we have a metal bicarbonate which reacts with an acid, our products are going to be carbon dioxide and water. We also will have a salt product. So let's show how this forms. We have potassium bicarbonate K H C 03, which is a solid or insoluble ionic compounds or salt, which reacts with our acid hydrochloric acid, which is aqueous because it's soluble in water on our reaction or on our product side. Rather, we have our first product, carbon dioxide, which is a gas, our second product, liquid water and then our third product, our salt where our salt is going to be formed from our metal, from our ionic compound or from our ionic salts. So that would be potassium because based on its position in group one, a potassium is a metal and our metal is going to be be bonded to our anion of our acid where we stated that we have our C L minus chloride anion. So we have potassium bonded to C L our anion here. And because our plus one charge of potassium is neutralized by the minus one charge on our chloride anion, we just have K C L. And this is going to be based on our soluble rules for chlorides, a soluble chloride. So we give it the label A Q for aqueous. Now, we need to make sure that this is a balanced equation. So we're going to compare our atoms on our reactant side, potassium hydrogen, carbon oxygen and chlorine to the same atoms on the product side, counting up our number of potassium. On the reactant side, we have a total of one mole for hydrogen. On the reactant side, we have two for carbon. On the reactant side, we have a total of one for oxygen on the reactant side given by the subscript three in our potassium bicarbonate. Next to oxygen, we have three total oxygens. And for chlorine on the reactant side, we have one. Now comparing our product side on our product side, we also have one mole of potassium. For hydrogen. On our product side, we have a total of two given by the subscript two. In our product water. For carbon. On the product side, we have a total of one mole for oxygen. On the product side given by the subscript two and carbon dioxide next to oxygen. And the one oxygen in our second product water, we have a total of three moles of oxygen on the product side. And then as far as chlorine, we have also just one mole of that on the product side. And based on our two sides of our reaction, our reactant side and our product side, we have the same number of atoms. So we can confirm that we have a balanced equation as written. So no coefficients are needed to be added. So our final answer is going to be our full equation between potassium bicarbonate and hydrochloric acid which produces carbon dioxide, water and potassium chloride as our products. So this entire equation is our balanced reaction and our final answer. I hope this made sense and let us know if you have any questions.

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Complete and balance the equation for each of the following reactions:c. HCl (aq) + NaHCO₃(s) →

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