Element M is prepared industrially by a two-step procedure according to the following (unbalanced) equations:
Assume that 0.855 g of M₂O₃ is submitted to the reaction sequence. When the HCl produced in step (2) is dissolved in water and titrated with 0.511 M NaOH, 144.2 mL of the NaOH solution is required to neutralize the HCl. (a) Balance both equations.

Question

Element M is prepared industrially by a two-step procedure according to the following (unbalanced) equations:

Assume that 0.855 g of M₂O₃ is submitted to the reaction sequence. When the HCl produced in step (2) is dissolved in water and titrated with 0.511 M NaOH, 144.2 mL of the NaOH solution is required to neutralize the HCl. (a) Balance both equations.

Accepted Answer

Hello everyone in this question asked to balance the following equations by adding the correct coefficients. We'll see you out A ceo and it's gonna yell, See you Plus c. 0. 2. We have copper, oxygen and carbon on both sides. We have one copper over here To oxygen and one carbon. one copper over here, two oxygen. And while carbon everything is balanced, we can add the coefficients one, one, one in one, And for any three and Plus H two L. And this yields N A O H plus NH three. We have sodium, nitrogen, hydrogen and oxygen on both sides. We have three sodium over here, one nitrogen to hydrogen and one oxygen. And we have one sodium over here, one nitrogen for hydrogen and one oxygen since we have three sodium on the reactive side and only one on the product side, we can multiply by three over here and put a three in front of any O. H. So now we're going to have three plus three for a total of six hydrogen for the most part. Factory over here, we're gonna add a three in front of H 20 And get six hydrogen over here. And now we have three oxygen over here And three over here. So we're going to put a one here, three here, three here and one here and for S. T. O. To let's go to building S. E. Oh three. Love selenium and oxygen on both sides. We have one selenium on the reactive side and for oxygen On the product side we have one selenium and three oxygen Since we have four auction on the reactive side and three on the product side, We need to find the lowest common multiple, which is going to be 12. I hope you must have about four over here, We're Gonna put a four in front of SC. So now we're gonna have four selenium. So if you put a forefront of S C. 02, we're gonna have four selenium over here, We're gonna have eight Plus two for a total of 10 auction over here. And if you put a two in front of 02, we're gonna have 12 oxygen. So we're gonna put a four here. A two here and a four here. Thanks for watching my video. And I hope it was helpful.

Verified Solution

Key Concepts

Stoichiometry

Balancing Chemical Equations

Acid-Base Neutralization