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Degger [83]
3 years ago
6

Type the correct answer in the box. Express your answer to two significant figures. An industrial vat contains 650 grams of soli

d lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with excess hydrochloric acid. This is the equation of the reaction: 2HCl + Pb(NO3)2 → 2HNO3 + PbCl2. What is the percent yield of lead(II) chloride?
Chemistry
1 answer:
SashulF [63]3 years ago
4 0

88.98 %

The Balance Chemical Equation is as follow,

                                  2 HCl + Pb(NO₃)₂     →    2 HNO₃ + PbCl₂

According to equation,

          331.2 g (1 mole) Pb(NO₃)₂ produces  =  278.1 g (1 mole) PbCl₂

So,

                 870 g of Pb(NO₃)₂ will produce  =  X g of PbCl₂

Solving for X,

                     X  =  (870 g × 278.1 g) ÷ 331.2 g

                     X  =  730.5 g of PbCl₂

Therefore,

               Theoretical Yield =  730.5 g

Also as given,

                Actual Yield =  650 g

So using following formula for percentage yield,

                        %age Yield  =  (Actual Yield / Theoretical Yield) × 100

Putting values,

                        %age Yield  =  (650 g / 730.5 g) × 100

                        %age Yield  =  88.98 %

Brianliest please and thank you.

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What is the volume of 2.5 moles of nitrogen gas (N2)<br>at standard temperature and pressure (STP)?​
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Answer:

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Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.
zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation  </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:  

  • Boyle's law at constant T, P = 1 / V  
  • Charles's law, at constant P, V = T  
  • Avogadro's law, at constant P and T, V = n  

So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

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T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm

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V=8.3 L

P=1.8 atm

n=5 mol

\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC

  • Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol

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