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WARRIOR [948]
3 years ago
10

Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O

2 What volume of hydrogen gas, as measured at P = 1.00 atm and T = 273.15 K, is required to react with 10.5 g of linolenic acid in this reaction?
Chemistry
1 answer:
Ludmilka [50]3 years ago
4 0

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g  = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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