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denpristay [2]
3 years ago
6

Calculate the theoretical yield for the amount of sodium carbonate produced as a result of this chemical reaction. Record your f

inal answer in the above data table.
Calculate the percent yield using your theoretical yield and the amount of sodium carbonate that was actually recovered. Record your final answer in the above data table.

How did your calculated mass of sodium carbonate compare with the actual mass you obtained from the experiment? If the two masses are different, suggest reasons for the difference.

Predict the amount of water and carbon dioxide that was produced as a result of this reaction.


1.

Mass of crucible

5.26 g

2.

Mass of crucible & NaHCO3

8.27 g

3.

Mass of NaHCO3 (2-1)

3.01g

4.

Theoretical Yield of Na2CO3
(Use the amount calculated in #3 as your starting amount)

g

5.

Mass of crucible & Na2CO3

7.13 g

6.

Mass of Na2CO3 – Actual Yield (5-1)

1.87 g

7.

% Yield = actual yield x 100
theoretical yield

%
Chemistry
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

98.6%

Explanation:

First we put down the equation representing the decomposition of sodium bicarbonate.

2NaHCO3(s) ----> Na2CO3(s) + H2O(l) + CO2(g)

We can see from the data provided that the mass of sodium bicarbonate reacted is 3.01 g.

The number of moles of sodium bicarbonate reacted= mass of sodium bicarbonate/ molar mass of sodium bicarbonate

Molar mass of sodium bicarbonate= 84.007 g/mol

Number of moles of sodium bicarbonate=3.01 g/ 84.007 g/mol

Number of moles of sodium bicarbonate= 0.0358 moles

From the balanced reaction equation,

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

Hence 0.0358 moles of sodium bicarbonate yields 0.0358/2 = 0.0179 moles of sodium carbonate

Theoretical yield of sodium carbonate = number of moles of sodium carbonate × molar mass of sodium carbonate

Molar mass of sodium carbonate= 105.9888 g/mol

Theoretical yield of sodium carbonate= 0.0179 moles × 105.9888 g/mol

Theoretical yield of sodium carbonate= 1.897 g

Actual yield of sodium carbonate= 1.87 g

%yield of sodium carbonate= 1.87/1.897 ×100

%yield of sodium carbonate= 98.6%

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Rust (Fe2O3. 4H2O) is formed when iron interacts slowly with oxygen and water. Mass of Fe in grams is 2.18 x 10⁴ g.

<h3>What is the explanation?</h3>

There are 2 moles of Fe atoms in 1 mole of Fe2O3-4H2O. The number of moles of Fe atoms in 45.2 kg rust is shown below.

Moles of Fe = 195.01 mol Fe₂O₃.4H₂O ( \frac{2 mol Fe }{1 mol Fe₂O₃.4H₂O} )

Moles of Fe = 390.02 mol Fe

Multiply the calculated number of moles of iron, Fe, by its molar mass which is 55.85 \frac{g}{mol}

Mass of Fe = 390.02 mol Fe ( \frac{55.85 g Fe}{mol Fe} )

Mass of Fe = 2.18 x 10⁴ g Fe

Avogadro's number (6.022 x 1023) of molecules (or formula units) make up one mole of a substance (ionic compound). The mass of 1 mole of a chemical is indicated by its molar mass. It provides you with the amount of grams per mole of a substance, to put it another way.

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1 year ago
What is the main process that happens in the mantle?
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When 125.0 g of ethylene (C2H4) burns in 60.0 grams of oxygen to give carbon dioxide and water, how many grams of CO2 are formed
gizmo_the_mogwai [7]

Answer:

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Solution:

The Balance Chemical Reaction is as follow,

                             C₂H₂  +  5/2 O₂    →    2 CO₂  +  H₂O

Or,

                             2 C₂H₂  +  5 O₂    →    4 CO₂  +  2 H₂O    -------  (1)

Step 1: Find out the limiting reagent as;

According to Equation 1,

            56.1 g (2 mole) C₂H₂ reacts with  =  160 g (5 moles) of O₂

So,

                  125 g of C₂H₂ will react with  =  X g of O₂

Solving for X,

                      X =  (125 g × 160 g) ÷ 56.1 g

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It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of CO₂ produced as;

According to Equation 1,

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So,

                  60.0 g of O₂ will produce  =  X g of CO₂

Solving for X,

                      X =  (60.0 g × 176 g) ÷ 160 g

                      X =  66 g of CO₂

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