Chemistry, 27.03.2020 06:16 brcribb

Problem PageQuestion Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.6 g of butane is mixed with 7.65 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

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Answer from: loverohio001

6.47 grams is the maximum mass of carbon dioxide that could be produced by the chemical reaction.

Explanation:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

Moles of butane = $\frac{4.6 g}{58 g/mol}=0.07931 mol$

Moles of oxygen gas = $\frac{7.65 g}{32 g/mol}=0.239 mol$

According to reaction , 13 moles of oxygen gas reacts with 2 moles of butane,then 0.239 moles of oxygen gas will react with :

$\frac{2}{13}\times 0.239 mol=0.0368 mol$ of butane.

From this we can see that butane is present in an excess amount and oxygen gas is present in limited amount. Hence, oxygen gas is a limiting reagent.

Since, oxygen gas is a limiting reagent which means that amount of carbon dioxide gas produce will depend upon moles of oxygen gas.

According to reaction, 13 moles of oxygen gas gives 8 moles of carbon dioxide, then 0.239 moles of oxygen gas will give:

$\frac{8}{13}\times 0.239 mol=0.147 mol$

Mass of 0.1471 moles of carbon dioxide gas:

0.147 mol × 44.01 g/mol = 6.47 g

6.47 grams is the maximum mass of carbon dioxide that could be produced by the chemical reaction.

Answer from: Quest

the ratio of the conjugate base to the acid required to attain the desired ph.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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Problem PageQuestion Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide...

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