Question

ch. 10,
chemical reactions

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placeholder for further questions. \[\large{\cancel{\color{red}{\text{PLEASE DO NOT ANSWER}}}}\] Since everybody decided to bypass the "Please don't answer as this is a placeholder for further questions" part, I have closed the question.

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SET ONE (original link: http://prntscr.com/bzd34t )

1.15.00 grams of iron are burned in 15.00 grams of oxygen to produce an iron (III) oxide compound. How many grams of iron (III) oxide are produced?

2.In photosynthesis, plants convert carbon dioxide and water into glucose (C6H12O6) according to the reaction (arrow is sunlight):
6CO2(g) + 5H2O(l) → 6O2(g) + C6H12O6
Suppose you determine that a particular plant consumes 37.8g of CO2 in one week. Assuming that there is more than enough water present to react with all of the CO2, what mass of glucose (in grams) can the plant synthesize from the CO2?

3.Given the following reaction: Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2]
a) How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?
b) What is the mass of NaBr that will be produced from 42.7 g of AgBr?

4.From the reaction: B2H6 + O2 → HBO2 + H2O
a) What mass of O2 will be needed to burn 36.1 g of B2H6?
b) How many moles of water are produced from 19.2 g of B2H6?

5.Calculate the mass (in kg) of water produced by the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasoline is 0.79 g/mL.

Answer 1

SET TWO

1) When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed
a) Write the balanced equation for previously stated reaction:
CuCl\(_{2}\) + NaNO\(_{3}\) → Cu(NO\(_{3}\))\(_{2}\) + NaCl
b) If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed?
c) What is the limiting reagent for the reaction in #2/(b)?
d) How many grams of copper (II) nitrate is formed?
e) How much of the excess reagent is left over in this reaction?

2) When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed.
a) Balance the following equation:
Pb(NO\(_{3}\))\(_{2}\) (aq) + NaI (aq) → PbI\(_{2}\)w (s) + NaNO\(_{3}\) (aq)
b) If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
c) What is the limiting reagent in the reaction described in this problem?
d) How many grams of lead (II) iodide is formed?
e) How much of the non-limiting reagent will be left over from the reaction in this problem?

Answer 2

You're posting too many questions at once, it will get confusing in the answers.

They're, more or less, the same stoichiometry problem.

1. Write reaction, balance
2. Convert to moles
3. Find limiting reactant
4. Use limiting reactant to find moles of product (or whatever they ask for)
5. Convert moles of product to grams

I'll do the first as an example.

1.15.00 grams of iron are burned in 15.00 grams of oxygen to produce an iron (III) oxide compound. How many grams of iron (III) oxide are produced?

1. \(\sf 4~Fe + 3~O_2 \rightarrow 2~Fe_2O_3\)

2. Find moles

\(\sf moles_{Fe}=\dfrac{mass}{Molar~mass}=\dfrac{15~g}{55.85~g/mol}=0.269~mol\)

\(\sf moles_{O_2}=\dfrac{15 ~g}{32~g/mol}=0.469~mol\)

3. Find limiting reactant

We need to divide the moles by the coefficients in the reaction

For Fe: \(\sf \dfrac{moles_{Fe}}{coefficient}=\dfrac{0.269~moles}{4}=0.0673~mol\)

For \(O_2\): \(\sf \dfrac{moles_{O_2}}{coefficient}=\dfrac{0.469~mo}{3}=0.156~mol\)

We have less Fe available for the reaction so Fe is the limiting reactant.

4. Find moles of \(Fe_2O_3\). We write ratios of the moles of whats of interest using their coefficients.

\(\sf \dfrac{moles_{Fe}}{4}=\dfrac{moles_{Fe_2O_3}}{2}\)

Solve for moles of Fe2O3: \(\sf moles_{Fe_2O_3}=2*\dfrac{moles_{Fe}}{4}=\dfrac{moles_{Fe}}{2}\)

Plug in moles from 2. (not from 3.)

\(\sf moles_{Fe_2O_3}=\dfrac{0.269~moles}{2}=0.1345~moles\)

5. \(\sf mass =moles*Molar~mass=0.1345~moles*159.69 ~g/mol=21.478~g\)

So 21.5 g of iron(III) oxide are made

Answer 3

@aaronq On the topic of too many questions, I did say "please do not answer" in the first line... because this is a placeholder for further questions...

which may or may not come from these ._.

Answer 4

But thanks for showing me the answers so that I can check them, I guess.

Answer 5

So you didn't want people to answer the questions? Then why post them at all?

Also, i didn't write the answers to all your questions, i wrote the solution for the first problem only.

Answer 6

\(\color{blue}{\text{Originally Posted by}}\) @aaronq
So you didn't want people to answer the questions? Then why post them at all?

Also, i didn't write the answers to all your questions, i wrote the solution for the first problem only.
\(\color{blue}{\text{End of Quote}}\)

\(\color{steelblue}{\text{Originally Posted by}}\) @/kittiwitti1
"this is a placeholder for further questions...

which may or may not come from these ._."
\(\color{steelblue}{\text{End of Quote}}\)

Answer 7

4.From the reaction: B2H6 + O2 → HBO2 + H2O

is this reaction balanced? if so how do you know

a) What mass of O2 will be needed to burn 36.1 g of B2H6?
b) How many moles of water are produced from 19.2 g of B2H6?

Answer 8

x_x

Answer 9

i never answered anything i'm asking you a question

Answer 10

idk I didn't get there yet

Answer 11

hence the usage of "placeholder" until I actually run into an issue ; - ;

Answer 12

I closed it because this is wasting everyone's time. I did say "don't answer" pretty clearly in the top half of the question.

Answer 13

start with one question first then give your approach to that.

Answer 14

& adding on top of that, the site keeps glitching every comment I post so I have to exit and re-enter the thread -_-