Chemistry, 26.05.2020 14:57 zalayjahan4793

What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 l of hydrogen gas

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Answer from: rebecabosca

Complete Question

What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 L of hydrogen gas at STP ?

The volume of butane produced is $V_b = 3.54 \ L$

Explanation:

From the question we are told that

The mass of carbon is $m_c = 13.45 \ g$

The volume of hydrogen is $V_h = 17.65\ L$

The number of moles of carbon is mathematically evaluated as

$n_c = \frac{m_c}{M_c}$

Where M_c is the molar mass of carbon which is a constant with value $M_c = 12\ g/mol$

$n_c = \frac{13.45}{12}$

n_c = 1.12

Now since the volume of the hydrogen is measured at standard temperature and pressure

Hence the number of moles of hydrogen is

$n_h = \frac{V_h }{22.4}$

Where 22.4 is a constant

$n_h = \frac{17.65}{22.4}$

n_h = 0.79

Comparing $n_h \ and\ n_c$ we see that hydrogen is the limiting reactant

because n_h is greater than n_c

The chemical equation for this reaction is

$4C_{(s)} + 5H_2-{(g)} ---- C_5 H_10_{(g)}$

looking at chemical equation we see that

5 moles of hydrogen gas reacts with 4 moles of carbon to produce 1 mole of butane

This implies that

0.79 moles react with 1.12 moles of carbon to produce x moles of butane

Therefore

$x = \frac{0.79}{5}$

$x = 0.158 \ moles$

Now since this reaction is carried out at standard temperature and pressure the volume of butane produced will be

V_b = 0.158 * 22.4

$V_b = 3.54 \ L$

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

Answer from: Quest

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.

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What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 l of hydrogen...

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