Balance each of the following chemical equations:
d. Al(s) + HCl(aq) → H₂(g) + AlCl₃(aq)

Question

Balance each of the following chemical equations:

d. Al(s) + HCl(aq) → H₂(g) + AlCl₃(aq)

Accepted Answer

Welcome back, everyone. We need to give the balanced chemical equation for the following reaction where solid gallium reacts with nitric acid to produce gallium, three nitrate and hydrogen gas as products. We're going to notice that gallium takes the place of hydrogen in our nitric acid to form our gallium three nitrate product. So we're going to recognize this as a single displacement reaction. Now, we're going to need to compare our reactants to our products. So beginning with listing out our atoms, we have gallium present in our reaction, hydrogen nitrogen and oxygen on both sides of our equation. We're going to recall that we want to balance out non hydrogen and non oxygen atoms first. So counting up our number of gallium on the reactant side, we have one mole of gallium, counting up hydrogen, we have one mole of hydrogen. For nitrogen, we would count one on the reactant side. And then for oxygen on the reactant side given by the subscript of three in nitric acid, we have three oxygens on the reactant side. Next, for our product side, we have one mole of gallium given by our gallium three nitrate product in hydrogen. On the product side, we would count from our subscript of two in hydrogen gas, a total of two hydrogens on the product side. For nitrogen, on the product side, given by our product gallium three nitrate. Outside of our parentheses, we have a subscript of three which would attribute to a total of three nitrogens on the product side. And then for oxygen on the product side and gallium three nitrate, our nitrate is in parentheses and we have of subs a subscript sorry of three next to oxygen. And on the outside of the parentheses, we have a subscript of three. So multiplying those subscripts of three, we would have a total of nine oxygens on the product side. And as we say it, we balance nonoxynol first. So observing that our ngen are not balanced, we're going to place a coefficient of three in front of our nitric acid reactant so that we now have three moles of nitrogen on the reactant side, meaning that it's balanced with the three moles of nitrogen on the product side. And this will change our number of hydrogen on the reactant side to now three. And we'll also change our oxygen on the reactant side where our coefficient of three is multiplied by a subscript of three next to oxygen in nitric acid. So we have a total of nine moles of oxygen on the reactant side. So now we need to bounce hydrogen And observing that we have three moles of hydrogen on the reactant side and only two on the product side, we're going to need to utilize a fraction so that we place 3/2 as our coefficient in front of our hydrogen gas on the product side. However, we want to have as many whole numbers as possible when having a balanced equation. So we're going to take the entire equation and multiply by two. This will then allow us to cancel out our denominator of two and three halves with our two. That we're multiplying the entire equation by which will now give us the following equation in which we now have two moles of solid gallium Reacting with six moles of nitric acid. When we have two, multiplied by a coefficient of three from our original equation, then this will yield two moles of gallium, three nitrate and just making more room for the rest of our equation, we would have this added to three moles of hydrogen gas as our second product. So now we can correct our final equation so that we can count a total of two moles of gallium on both sides of our equation. We would now count for hydrogen on the reactant side, a total of Six moles of hydrogen on the reactant side along with six on our product side, where we see our coefficient of three is multiplied by the subscript of two in hydrogen gas. And for nitrogen, our nitrogens will now be a total of three on both sides of our equation, which we already had. And then for oxygen, we would have 18 moles of oxygen. Now, on the reactant side to match with the 18 moles of oxygen on the product side, which we get by taking our Coefficient of two in gallium, three nitrate and multiplying it by our subscript of three inside of our parentheses of our nitrate portion of the compound. And then that would give us a total of six, which is then multiplied by the subscript of three Outside of our parentheses of nitrate, which would give us a total of 18 oxygens. So now we would confirm that our final answer is going to be two moles of gallium reacting with six moles of nitric acid to produce two moles of gallium, three nitrate and three moles of hydrogen gas as our correct balanced equation for our given reaction. So I hope this made sense and let us know if you have any questions.

Balance each of the following chemical equations:
d. Al(s) + HCl(aq) → H₂(g) + AlCl₃(aq)

Verified video answer for a similar problem:

Key Concepts

Chemical Equation

Balancing Chemical Equations

Stoichiometry

Balance each of the following chemical equations:d. Al(s) + HCl(aq) → H2(g) + AlCl3(aq)

Verified video answer for a similar problem:

Key Concepts

Chemical Equation

Balancing Chemical Equations

Stoichiometry

Balance each of the following chemical equations:
d. Al(s) + HCl(aq) → H₂(g) + AlCl₃(aq)