Balance each of the following chemical equations:
c. Sb₂S₃(s) + HCl(aq) → SbCl₃(aq) + H₂S(g)

Question

Balance each of the following chemical equations:

c. Sb₂S₃(s) + HCl(aq) → SbCl₃(aq) + H₂S(g)

Accepted Answer

Welcome back, everyone. We need to write the balanced chemical equation for the following reaction where we have antimony three oxide solid reacting with hydrotic acid to produce antimony, three iodide and liquid water as products notice that antimony takes the place of hydrogen to form our antimony three iodide product. So we would recognize this as a single displacement reaction. Now, we're going to need to compare each side of our equation. So we're going to compare our reactant side to our product side. On our reactant side, we have the atoms antimony, oxygen, hydrogen and iodine. So we have the same on our product side where on our reactant side, we would count a total given by our subscript of two and antimony three oxide. We have two antimony on our reactant side. As far as oxygen given by the subscript of three and antimony three oxide, we have three oxygens. And then for hydrogen on the reactant side in our reactant hydrotic acid, we have just one mole of hydrogen and we also have one mole of iodine. Now, on our product side, counting our number of antimony in our product Antimony three Iodide, we have just one of our antimony for oxygen. On the product side, in our product water, we have just one of our water, sorry, just one of our oxygens. And then as far as hydrogen on the product side, given by our subscript of two in our product water, we have two hydrogens. And as far as iodine on the product side, given by our subscript of three in antimony, three iodide, we have three iodines. So we're going to begin by balancing out non oxygen and non hydrogen atoms first. So let's start with antimony. We would observe that we have two moles of antimony on the reactant side and just one on the product side. So we can mend this by placing a coefficient of two in front of our antimony three iodide in which we would now have a total of two moles of antimony on the product side. And this will also this coefficient of two will also change our number of iodine on the product side to now being multiplied by the subscript of three. So that we have a total of six moles of iodine on the product side. So next, let's balance our atom of iodine in which we have six on the product side and just one on the reactant side. So we can place a coefficient of six in front of our reactant hydrotic acid, which is now going to change our hydrogen on the reactant side to now six. And our iodine on the reactant side to now six to match with our iodine on the product side. So next, let's go ahead and balance hydrogen where we have six on the reactant side and two on the product side. So in order to balance hydrogen, we're going to go ahead and place a coefficient of three in front of our product, liquid water to which we would now multiply by a subscript of two in water, which would give us the total of six moles of hydrogen on the product side to match with the six on the reactant side. And adding that coefficient of three changes our oxygen on the product side to now a total of three. And so now with these changes, we have a fully balanced equation. So our final answer is going to be that one mole of antimony, three oxide solid reacts with six moles of hydro ionic acid to produce two moles of antimony, three iodide and three moles of water. So just so this is visible, gonna move this over a bit. And this entire equation with our coefficients is our final answer as the correct balanced chemical equation. So I hope this made sense and let us know if you have any questions.

Balance each of the following chemical equations:
c. Sb₂S₃(s) + HCl(aq) → SbCl₃(aq) + H₂S(g)

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Balance each of the following chemical equations:
c. Sb₂S₃(s) + HCl(aq) → SbCl₃(aq) + H₂S(g)