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spin [16.1K]
3 years ago
12

When lead nitrate reacts whith sodium iodide, sodium nitrate and lead iodide are formed if i start with 25.0 grams of lead nitra

te and 15.0 grams of sodium iodide, find the percent yield if this reaction forms 6.0 g of sodium nitrate during the experiment
Pb(NO3)2+ 2NaL--> PbL2+ 2NaNO3
Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
4 0

So you find the moles of the lead nitrate by doing mass/Mr so 25/331.2 which gives you 0.075

Due to stoichiometry in the reaction we see 1:2 ratio so we do 0.075x2 gives 0.15 moles of NaNO3

Then mass: moles x Mr so 0.15 x 85= 12.75g is your theoretical

So % yield is actual/ theoretical x 100

Therefore 6.0g/12.75 g x 100 = 47%

Not sure if it’s correct hope it helps!

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3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O
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24.25 moles of NO can be produced using 97 moles of HNO3.

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3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒  8 HNO3⇔2 NO

⇒ 1 HNO3⇔ 1/4 NO

This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , \frac{1}{4}  *97 moles of NO can be made,

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2 years ago
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Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution i
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Answer:

y_{C_6H_{14}}=0.92

Explanation:

Hello,

At first, we compute liquid-phase molar fractions:

n_{C_6H_{14}}=463.96 g*\frac{1mol}{86g} =5.3949molC_6H_{14}\\n_{C_7H_{16}}=667.71 g*\frac{1mol}{100g} =6.6771molC_7H_{16}\\x_{C_6H_{14}}=\frac{5.3949}{5.3949+6.6771} =0.447\\x_{C_7H_{16}}=1-x_{C_6H_{14}}=0.553

Now, by means of the fugacity concept, for hexane, for instance, we have:

f_{C_6H_{14}}^V=f_{C_6H_{14}}^L\\y_{C_6H_{14}}p_T=x_{C_6H_{14}}p_{C_6H_{14}}

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

p_T=x_{C_6H_{14}}p_{C_6H_{14}}+x_{C_7H_{16}}p_{C_7H_{16}}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

y_{C_6H_{14}}=\frac{x_{C_6H_{14}}p_{C_6H_{14}}}{p_T}=\frac{0.447*0.198946 atm}{0.096622atm} \\y_{C_6H_{14}}=0.92

Best regards.

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3 years ago
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