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2 years ago

How many moles of PbCl2 are produced if 14 moles of AIC13 are consumed?

Chemistry

2 answers:

swat322 years ago

5 0

Answer:

14 moles AlCl3 x 3 moles PbCl2/2 moles AlCl3 = 21 moles of Pb(NO3)2

Explanation:

Using dimensional analysis and the mole ratios from the balanced equation, we can calculate moles of PbCl2/2 formed from 14 moles of AlCl3...

I hope this helps and if I am wrong I am sorry but i am positive that your answer.

alexandr1967 [171]2 years ago

5 0

Answer:

21 moles of PbCl2 are produced

Explanation:

In order to produce PbCl2 from AlCl3 some reaction is needed. Assuming all the chlorine in the reactants is present in the AlCl3 and all the chlorine in the products is present in the PbCl2, then:

Pb compound + 2 AlCl3 -> 3 PbCl2 + Al compound

The aforementioned chlorine assumption is needed to find the stoichiometric coefficients of the balanced equation. So, for every 2 moles of AlCl3 that react, 3 moles of PbCl2 are formed, then for 14 moles of AlCl3:

2 moles of AlCl3/14 moles of AlCl3 = 3 moles of PbCl2/x moles of PbCl2

x = 3*14/2

x = 21 moles of PbCl2 formed

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2 years ago

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2 years ago

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

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2 years ago

Find the time Δt it takes the magnetic field to drop to zero. Express your answer in terms of some or all of the quantities a, B

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2 years ago

Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo

Hitman42 [59]

Answer:

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Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

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Now this equation has a form y = mx + b where

y = lnP

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m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a software as excel or your calculator.

T ( K) 1/T ln P

377 0.002653 5.9915

384 0.002604 6.2115

390 0.002564 6.3969

395 0.002532 6.5511

398 0.002513 6.6333

The best line has a fit:

y = -4609.5 x + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

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Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0

2 years ago