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bekas [8.4K]
3 years ago
8

Solid sodium reacts violently with water producing heat, hydrogen gas and sodium hydroxide. How many molecules of hydrogen gas a

re produced when 65.4 g of sodium are added to water?
2 Na(s) +2 H2O(l) -- > 2 NaOH(aq) + H2(g)

Show all work.

Chemistry
2 answers:
nignag [31]3 years ago
4 0
Find the number of moles of sodium you have: 
<span>n = m/M where m is your 20g of sodium and M is 22.99 g/mol. </span>

<span>Look at the stoichiometry of the equation - it's 2:2 when you are producing NaOH. So if you took 1 mole of Na, it'd produce 1 mole of NaOH (as the ratio is equal). </span>

<span>That means that your moles of sodium is equal to the moles of NaOH produced. Use the molar mass of NaOH - which is 39.998 g/mol along with your calculated number of moles to get the mass (the formula rearranges to m = nM). </span>

<span>This figure is the theoretical yield - what you would get if every last mole of sodium was converted into NaOH. </span>

<span>What you get in practice is the experminetal yield, and the percentage yield is the experimental yield divided by the theoretical yield - and then multiplied by 100%.</span>
sergeinik [125]3 years ago
3 0
Hope it helped you.

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Which of the following reactions have a positive ΔSrxn? Check all that apply.
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Answer:

The reactions that have a <em>positive ΔS rxn </em>are the first and the fourth choices:

  • <em>2A(g) + B(s) → 3C(g)</em>

  • <em>2A(g) + 2B(g) → 5C(g)</em>

Explanation:

<em>ΔS rxn </em>is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn  means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

  • <em><u>2A(g) + B(s) → 3C(g)</u></em>

        1)  The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

        2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: <em>ΔS rxn is positive.</em>

  • <em><u>2A(g) + B(g) → C(g)</u></em>

Using the same reasoning, 3 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>A(g) + B(g) → C(g)</em></u>

Again, 2 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>2A(g) + 2B(g) → 5C(g)</em></u>

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: <em>ΔS rxn is positive.</em>

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