Write a balanced molecular equation describing each of the...

Write a balanced molecular equation describing each of the following chemical reactions. (Use the lowest possible whole number coefficients. Include states-of-matter under the given conditions in your answer.) (a) Solid beryllium carbonate is heated and decomposes to solid beryllium oxide and carbon dioxide gas. (b) Liquid octane, C8H18, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor. (c) Aqueous solutions of nickel(II) nitrate and lithium hydroxide react to produce solid nickel(II) hydroxide and aqueous lithium nitrate. (d) Water vapor reacts with rubidium metal to produce solid rubidium hydroxide and hydrogen gas.

Transcript

00:01 Hi, so we are to write the balanced molecular equation for the given reactions here.

00:09 So for reaction a, we have solid beryllium carbonate.

00:14 So beryllium carbonate, that's peco3.

00:19 And this will decompose to the products beryllium oxide, that's po, and carbon dioxide, that's co2.

00:31 Now the states of matter for the compounds, we have solid beryllium carbonate.

00:37 So this is s.

00:39 This is solid beryllium oxide and of course carbon dioxide is gaseous.

00:44 So you write g.

00:46 Now we need to balance this equation.

00:49 Let's count the number of elements on both sides of the equation.

00:52 Beryllium we have 1.

00:54 Carbon we have 1.

00:55 Oxygen we have 3.

00:57 This is on the reactant side.

00:58 On the product side, beryllium we have 1.

01:02 Carbon we also have one oxygen we have one plus two so that's three as well so we now have equal number of elements in both sides of the equation therefore this equation is already balanced let me just i'm sorry about that i'm gonna box this and see balance equation for the first reaction now for the second reaction we have octane that's ch18 this is liquid and we have oxygen gas this is combustion reaction and the products are co2 and water vapor.

01:51 For.

01:51 So co2, that's gas, and also h2o, this is in gaseous space.

01:58 Now we need to balance this equation.

02:01 Since we have eight carbon atoms in the product's reactant side, and we only have one on the product side, that means we need to write eight in here, so that we'll have equal number of carbon on both sides of the equation.

02:14 Now let's balance the number of hydrogen.

02:16 We have 18 on the the reactant side and only 2 on the product side.

02:19 So that means we have to write here 9.

02:24 Now let's count the number of elements on both sides of the equation.

02:29 Carbon is 8, hydrogen is 18, oxygen is 2.

02:34 On the product side, we have carbon 8, hydrogen 9 times 2, that's also 18.

02:41 Oxygen we have 2 times 8 or 8 times 2 plus 9 multiplied by 1 oxygen is therefore equivalent to 25 atoms and since we have 2 on the reactant side we need to balance this and we'll do that by by writing 25 over 2 here.

03:11 But since the instruction for this problem is to use the lowest possible whole number coefficient, that means we have to multiply both sides of the equation by 2, so that we'll have a whole number coefficient for oxygen gas...

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