Write balanced equations for the following reactions:
(c) phospho...

Question

Write balanced equations for the following reactions:
(c) phosphorus pentoxide with water

Accepted Answer

welcome back everyone in this example, we need to give a bounce equation that represents the reaction between tungsten dioxide and water. So our first step is to write out our reaction and our compounds involved. So we have tungsten dioxide given as W. 02. And then it's reacting with H 20. Now recall that tungsten on our periodic table is considered a metal and it's bonded to our non metal here being oxygen. So we should recognize tungsten dioxide as a metal oxide. And we should recall that when we have a metal oxide reacting with water, we're going to form a metal hydroxide product and this would be our only product, which is why we should recognize that this is specifically going to be a synthesis reaction Because we have two agents Reacting together to Form one single product. Now, before we go ahead and write out our product, we want to make note of our charge on our tungsten because we should specifically recognize that tungsten on our product table is in our transition metal D block. So it's a transition metal. And so that means we need to figure out its oxidation number A K A H charge. Recall that oxidation is losing electrons and so losing electrons creates a charge. So we would recall that because tungsten is a transition metal, it can form various charges. So we're going to have to solve for what its oxidation number would be when it's bonded to these two oxygen atoms here. So we would say That if we set up an equation with regard to tungsten unknown oxidation number as X. And oxygen's oxidation number set equal to zero. We can go ahead and figure out what tungsten oxidation number would be. So we would have an equation where we have zero equal to X. And then we would say plus our oxidation number for oxygen we should recall is -2. So we'll place that in parentheses and we have two oxygen atoms because we have that subscript of two. So we would say -2 times two. Again, we have that minus two for oxygen's oxidation number because we recognize that oxygen is in Group six A. On our product table. And we recall that Adams in Group six a form a minus to an ion charge. And so solving for X here, so let's go ahead and actually move this to the right. We would say that X. Is equal to plus four. And so that means that tungsten therefore has a plus four oxidation number or caddy in charge. And so for our metal hydroxide product we would have our tungsten metal and then we have in parentheses hydroxide, which we recall is O. H. And because we are crossing the charges between our two re agents here, we recall that hydroxide has a minus one charge, which is why we just wrote tungsten alone as is. And then we just figured out that tungsten has a charge of plus four. So we would have a subscript of four on our metal hydroxide product. So this is our metal hydroxide product tungsten hydroxide. And now we need to make sure that we have a balanced equation. So we're going to compare the atoms on our reactant side to the atoms on our product side. So we have tungsten, we are comparing, we have hydrogen and we have oxygen and the same thing on our product side, tungsten, hydrogen and oxygen. So we can count a total of just one atom of tungsten on our reactant side, A total of three atoms of oxygen on our product side. Or sorry, on our reactant side, A total of just two hydrogen atoms on our reactant side. And now moving on to our product side, we have one atom of tungsten counting out our oxygen's because it's in the parentheses with the subscript four, we have four, oxygen's on the product side and then the same applies to our hydrogen. So we need to go ahead and alter our coefficients so that we have matching number of atoms on both the reactant and product side. And so we would say that if we place a coefficient of two in front of our water, that would change our high regions on our reactant side to now, a total of four because we would have two times our subsequent to there. And then we would also change our oxygen's from being three atoms of oxygen on the reacting side to now, a total of two plus another two here. And so that would give us four. And so that would actually balance all of our hydrogen, oxygen and tungsten atoms so we have a balanced reaction. And so for our final answer, that is going to be our bounced equation here where we have our metal oxide forming our metal hydroxide product. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Write balanced equations for the following reactions: (c) phosphorus pentoxide with water

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