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garik1379 [7]
3 years ago
12

Calculate the number of moles of O2 produced using the ideal gas law. Then, use this value to calculate the number of moles of h

ydrogen peroxide you began the experiment with. HINT: Use the balanced equation provided in the lab introduction.
Chemistry
1 answer:
aev [14]3 years ago
6 0

Answer:Given data from the lab collated as 291.15k for Temperature, 0.061L for volume and 1atm pressure.

Initial hydrogen peroxide solution is 5ml with a weigh % concentration 6.6%

n=0.0025moles of oxygen AND 0.0097moles of H2O2 were present

Explanation:

2H2O2(aq)-----2H2O(I)+2O2(g)

Recall the idea gas equation

PV=nRT

n=PV/RT---(1) solving equation (1) using the data in the answer section above.

n=0.0025moles of oxygen is produced.

Recall that for every 100ml of solution you get 6.6g of hydrogen peroxide.

Your 5ml will thus contain

5ml*6.6g H2O2/100ml=0.33g

no of moles if H2O2=0.33/34.015=0.0097moles

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3 years ago
Explain Rutherford’s gold foil experiment. What particle did he discover?
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3 years ago
How would you describe the image formed by a convex mirror? a. upright and larger than object c. inverted and larger than object
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3 0
3 years ago
Read 2 more answers
1: Write balanced complete ionic equation for
monitta

Answer 1 : The balanced complete ionic equation will be,

NaOH(aq)+HNO_3(aq)\rightarrow H_2O(l)+NaNO_3(aq)

Na^+(aq)+OH^-(aq)+H^+(aq)+NO_3^-(aq)\rightarrow H_2O(l)+Na^+(aq)+NO_3^-(aq)

By removing the spectator ion in this equation, we get the balanced ionic equation.

OH^-(aq)+H^+(aq)\rightarrow H_2O(l)

Answer 2 : The balanced net ionic equation will be,

2Na_3PO_4(aq)+3NiCl_2(aq)\rightarrow Ni_3(PO_4)_2(s)+6NaCl(aq)

6Na^+(aq)+2PO_4^{3-}(aq)+3Ni^+(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)

By removing the spectator ion in this equation, we get the balanced ionic equation.

2PO_4^{3-}(aq)+3Ni^+(aq)\rightarrow Ni_3(PO_4)_2(s)

Answer 3 : The balanced net ionic equation will be,

2Na_3PO_4(aq)+3NiCl_2(aq)\rightarrow Ni_3(PO_4)_2(s)+6NaCl(aq)

6Na^+(aq)+2PO_4^{3-}(aq)+3Ni^+(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)

By removing the spectator ion in this equation, we get the balanced ionic equation.

2PO_4^{3-}(aq)+3Ni^+(aq)\rightarrow Ni_3(PO_4)_2(s)

Balanced equations : Balanced equations are the equations in which the number of individual elements present on the reactant side must be equal to the number of individual elements present on the product side.

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7 0
3 years ago
Read 2 more answers
Excess oxygen gas (O2) reacts with 244g of Iron (Fe) to produce 332 g of Fe2O3. What is the percent yield?
Andru [333]

Answer:

95.15%

Explanation:

To calculate the percent yield, we need the following formula:

\% yield=\dfrac{actual yield}{theoreticalyield}\times100\%

Solving for the theoretical yied, you need to predict how much of the product will be produced if we USE up the given.

Our given is 224g of Fe and we to get the theoretical yield, we need to figure out how much product will Fe produce supposing that we use up all the reactant.

First thing we do is get the balance equation of this chemical reaction:

<u>4</u>Fe + <u>3</u>O₂ → <u>2</u>Fe₂O₃

We get the ratio between Fe and the the product, Fe₂O₃

\dfrac{4moles of Fe}{2molesofFe_{2}O_{3}}=\dfrac{2moles of Fe}{1moleofFe_{2}O_{3}}

This basically means that we need 2 moles of Fe to produce 1 mole of Fe₂O₃. We'll use this later.

Now we let's use our given:

We need to first convert our given to moles. To do this, we need to determine how many grams there are of the reactant for every mole. We need to first get the atomic mass of the elements involved in the substance:

          Iron(1)      

Fe  =   55.845(1) = 55.845g/mole(1)

Then we use this to convert grams to moles

244g\times\dfrac{1mole}{55.845g}=4.369moles

This means that there are 4.396moles of Fe in 244g of Fe.

This we will use to see how many moles of product we can produce given the moles of reactant by using the reactant:rproduct ratio.

4.369moles of Fe\times\dfrac{1moleofFe_{2}O_{3}}{2molesofFe}=2.185 moles of Fe_{2}O_{3}

So given 4.693 moles of Fe we can produce 2.185 moles of Fe₂O₃

The next step is to get how many grams of product there are given our calculation. We do this again by getting how many grams of Fe₂O₃ there are in 1 mole.

               Fe(2)              O(3)    

Fe₂O₃=55.845(2)  +  15.999(3)

         = 111.69        +   47.997      =159.687g/mol

We then use this to solve for how many grams of product there are in 2.185 moles.

2.185moles\times\dfrac{159.687g}{1mole}=348.92g

This is our theoretical yield 348.92g of Fe₂O₃.

We can finally use our percent yield equation. Our actual yield is given by the probelm, <u>332g of Fe₂O₃</u> and we solved for our theoretical yield which is <u>348.92g of Fe₂O₃</u>. We plug this in our formula and solve.

\%yield=\dfrac{actual yield}{theoreticalyield}\times100\%

\%yield=\dfrac{332gofFe_{2}O_{3}}{348gofFe_{2}O_{3}}\times100\%=95.15\%

So the answer is 95.15%

3 0
3 years ago
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