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Ksivusya [100]
3 years ago
8

1) How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7g Fe? Show your work.

Chemistry
2 answers:
klasskru [66]3 years ago
6 0
1.) is157.7 g
<span>moles Fe = 209.7 g/ 55.847 g/mol=3.75 
the ratio between Fe and CO is 2 : 3 
moles CO = 3.75 x 3 /2 =5.63 
mass CO = 5.63 mol x 28.01 g/mol=157.7 g

2.) is </span><span>1.06 moles 
</span>48.7/23 = 2.12 moles sodium 
<span>2.12/2 x 24 = 25.44dm^3 hydrogen = 1.06 moles </span>
1.06 X 6.02x10^23 = 1.204x10^24 molecules of hydrogen.

3.) is 91.8 
<span>8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced. </span><span>8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically. </span><span>% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 </span>
Travka [436]3 years ago
5 0
Mol Fe = 209.7/56 = 3.744 mol Fe
mol CO = 3.744 * 3/2 (mol ratio) = 5.616 mol CO
molecular weight of CO = 28
5.616 * 28 = 157.28 g CO needed

48.7/23 = 2.11 mol Na
2.11 * 1/2 = 1.055 mol H2
1.055 * 6.023*10^23 = 6.351 x 10^23 molecules of H2

8.3 * 2/2 = 8.3 mol H2O that could be produced
137.1/18 = 7.166 mol actually produced
7.616/8.3 * 100% = 91.75% produced (% yield)
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If the change of enthalpy of this reaction when proceeding left to right is +14 kcal, which chemical equation is correct?
bagirrra123 [75]

Answer:

N₂O₄ + 14 kcal ⇄ 2NO₂.

Explanation:

Since the sign of ΔH determines either the reaction is exothermic or endothermic:

+ve, the reaction is endothermic.

-ve, the reaction is exothermic.

∵ The change of enthalpy of this reaction when proceeding left to right is + 14 kcal (+ ve sign).

∴ The reaction is endothermic, the heat is a part of the reacatnts in the reaction.

So, the reaction is:

N₂O₄ + 14 kcal ⇄ 2NO₂.

7 0
3 years ago
What volume of O2 is required to react with CS2 to produce 4 liters of CO2?
Alenkasestr [34]

Answer:

12 L of O₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CS₂ + 3O₂ —> CO₂ + 2SO₂

From the balanced equation above,

3 L of O₂ reacted to produce 1 L of CO₂.

Finally, we shall determine the volume of O₂ required to produce 4 L of CO₂. This can be obtained as follow:

From the balanced equation above,

3 L of O₂ reacted to produce 1 L of CO₂.

Therefore, xL of O₂ will react to produce 4 L of CO₂ i.e

xL of O₂ = 3 × 4

xL of O₂ = 12 L

Thus, 12 L of O₂ is needed for the reaction.

3 0
3 years ago
Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat
Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

      Turnover number = K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}

     V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}

Therefore, moles of CO_{2} hydrolyzed is as follows.

Moles of CO_{2} hydrolyzed = \frac{Mass of CO_{2}}{Molar mass of CO_{2}}

                 = \frac{0.30}{44}

                 = 0.00682 moles

Now, moles of CO_{2} hydolyzed per second is calculated as follows.

Moles of CO_{2} hydolyzed per second = \frac{0.00682}{60}

             = 1.137 \times 10^{-4} moles/second = V_{max}

And,

Moles of enzyme = \frac{Mass}{\text{Molar mass}}

                       = \frac{10.0 \mu g}{30000}

                       = 3.33 \times 10^{-10} moles

Therefore, the value of K_{cat} is as follows.

    K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}

               = 0.3411 \times 10^{6} per second

               = 0.3411 \times 60 \times 10^{6} per minute

               = 20.466 \times 10^{6} per minute

Thus, we can conclude that the turnover number (K_{cat}) of carbonic anhydrase (in units of min^{-1}) is 20.466 \times 10^{6} per minute.

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