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3 years ago

For the reaction 4 FeCl2(aq) + 302(g) → 2Fe2O3(s) + 4Cl2(g), what volume of a

Chemistry

1 answer:

Darya [45]3 years ago

5 0

Answer:

$V=10.12mL$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to evidence the 4:3 mole ratio between oxygen and iron (II) chloride; thus, we can compute the moles of the latter that are consumed by the given molecules of the former:

$n_{FeCl_2}=4.32x10^{21}molec O_2*\frac{1molO_2}{6.022x10^{23}molec O_2} *\frac{4molFeCl_2}{3molO_2} \\\\n_{FeCl_2}=0.0095molFeCl_2$

Now, since we have a 0.945-M solution of this iron (II) chloride, the corresponding volume turns out to be:

$V=\frac{n_{FeCl_2}}{M}\\\\V=\frac{0.00956mol}{0.945mol/L}\\\\V=0.01012L\\\\V=10.12mL$

Best regards!

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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

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Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
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$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

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