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GuDViN [60]
2 years ago
7

Lab: Limiting Reactant and Percent Yield

Chemistry
2 answers:
creativ13 [48]2 years ago
7 0

Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.

Explanation :

While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.

Step 1 : first write the chemical reaction and then balanced the chemical equation.

C_2H_4+H_2\rightarrow C_2H_6

Step 2 : convert the given masses into the moles if mass of C_2H_4 is 10.5 g and molar mass of C_2H_4 is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.

\text{moles of }C_{2}H_{4}=\frac{10.5g}{28g/mole}=0.375moles

\text{moles of }H_{2}=\frac{0.40g}{2g/mole}=0.20moles

Step 3 : Now we have to determine the limiting reagent and excess reagent.

\text{ moles of }C_{2}H_{4}\text{ in excess}=0.375-0.20=0.175\text{moles}

Now we conclude that C_{2}H_{4} is the limiting reagent and hydrogen is an excess reagent.

Hypothesis :

Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete.  Limiting reagent limits the formation of products.

OverLord2011 [107]2 years ago
7 0

Answer:

If a substance is the limiting reactant, then . . .

it will limit the formation of products

because . . .

the reaction is presented at limited

Explanation:

just took the test

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  <u><em>calculation</em></u>

 The  amount  of  moles  is calculated  using  the Avogadro's  law   constant

That  is  1  moles =  6.02 × 10²³  formula units

                 ? moles = 3.34 × 10³⁴ formula  units

<em>by cross  multiplication</em>

= {1 mole  ×3.34 ×10³⁴  formula units ] / 6.02 × 10²³ formula units}  = 5.55 ×10¹⁰  moles


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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
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Explanation:

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                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

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           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

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                             = 309.8 \times 10^{5} Pa

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Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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