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netineya [11]
3 years ago
7

A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 986 s

for 1.00 L of this gas to effuse. Under identical experimental conditions it required 811 s for 1.00 L of chlorotrifluoromethane (CClF3) gas to effuse. Calculate the molar mass of the unknown gas . (Remember that the faster the rate of effusion, the shorter the time required for effusion; that is, rate and time are inversely proportional.)
Chemistry
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

The molar mass of the unknown gas is 154.4 g/mol

Explanation:

Step 1: Data given

It takes 986 seconds for an unknown gas to effuse

It takes 811 seconds for chlorotrifluoromethane (CClF3) gas to effuse

Molar mass of chlorotrifluoromethane (CClF3) = 104.46 g/mol

Step 2: Calculate the molar mass of the gas

t1/t2/ √(M1/M2)

⇒with t1 = the time needed for the unknown gas to effuse = 986 seconds

⇒with t2 = the time needed for CClF3 to effuse = 811 seconds

⇒with M1 = the molar mass of the unknown gas = TO BE DETERMINED

⇒with M2 = the molar mass of CClF3 = 104.46 g/mol

986/811 = √(M1/104.46)

(986/811)² = M1 / 104.46

M1 = 154.4 g/mol

The molar mass of the unknown gas is 154.4 g/mol

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Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

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