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sesenic [268]
3 years ago
9

For the reaction ? Ch3oh+? O2 →? Hco2h+? H2o , what is the maximum amount of hco2h (46.0254 g/mol) which could be formed from 6.

95 g of ch3oh (32.0419 g/mol) and 8.11 g of o2 (31.9988 g/mol)? Answer in units of g
Chemistry
1 answer:
saw5 [17]3 years ago
8 0

Answer:- 9.98 grams.

Solution:- The given balanced equation is:

CH_3OH+O_2\rightarrow HCO_2H+H_2O

From this equation, there is 1:1 mol ratio between methanol and formic acid. We start with given grams of methanol and convert them to moles. Then using mol ratio the moles of foramic acid are calculated and finally converted to grams on multiplying by it's molar mass.

The calculations could easily be done using dimensional analysis as:

6.95gCH_3OH(\frac{1molCH_3OH}{32.0419g})(\frac{1molHCO_2H}{1molCH_3OH})(\frac{46.0254gHCO_2H}{1molHCO_2H})

= 9.98gHCO_2H

Hence, 9.98 grams of formic acid can be formed from 6.95 grams of methanol.

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A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
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D. How the pull of gravity has changed.
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3 years ago
Dependent variable? ( what did we measore)?
lana66690 [7]
The thing u change.
6 0
2 years ago
A stone is tied to a 0.85-meter cord. it is swung in a circle at a constant rate of 6.0 m/s. what is the centripetal acceleratio
Elan Coil [88]
The correct answer is 42. I know this answer is right because i have already turned my assignment in and got it right.
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3 years ago
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A 32.8 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?
olasank [31]

Answer:

32.8g/mole

Explanation:

Given parameters:

Mass of sample of gas = 32.8g

Volume  = 22.4L

Unknown:

Molecular weight  = ?

Solution:

To solve this problem we must understand that at rtp;

             1 mole of gas occupies a volume of 22.4L

 

Number of mole of the gas  = 1 mole

 Now;

   Mass  = number of moles x molecular weight

    molecular weight  = \frac{mass}{molecular weight}    = \frac{32.8g}{1mole }      = 32.8g/mole

8 0
2 years ago
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