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Phantasy [73]
3 years ago
15

a 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. If we are able to determine that 1.9 moles of carbon dioxide esc

aped before the container could be sealed, what is the new volume of the container?​
Chemistry
2 answers:
myrzilka [38]3 years ago
8 0

Hello!

A 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. If we are able to determine that 1.9 moles of carbon dioxide escaped before the container could be sealed, what is the new volume of the container?​

V1 (initial volume) = 25.5 L

n1 (initial number of moles) = 3.5 mol

V2 (final volume) = ? (in L )

Note: there was a leak in the number of moles, so we have:

n2 (final number of moles) = 3.5 mol - 1.9 mol = 1.6 mol

<u><em>By Avogadro's Law it is known that the volume is directly proportional to the number of gas particles, that is, the larger the number of moles of gas, the greater its volume, on which we have the following relation:</em></u>

\boxed{\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}}

We apply the data to the formula, we have:

\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}

\dfrac{25.5}{3.5} = \dfrac{V_2}{1.6}

multiply the means by the extremes

3.5*V_2 = 25.5*1.6

3.5\:V_2 = 40.8

V_2 = \dfrac{40.8}{3.5}

V_2 = 11.657... \to \boxed{\boxed{V_2 \approx 11.66\:L}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:</em></u>

<u><em>The new volume of the container is approximately 11.66 liters</em></u>

________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

kap26 [50]3 years ago
4 0

Answer:

11.66 L.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If P and T are constant, and have different values of n and V:

<em>(V₁n₂) = (V₂n₁).</em>

V₁ = 25.5 L, n₁ = 3.5 mol.

V₂ = ??? L, n₂ = 3.5 mol - 1.9 mol = 1.6 mol.

<em>∴ V₂ = (V₁n₂)/(n₁)</em> = (25.5 L)(1.6 mol)/(3.5 mol) =<em> 11.66 L.</em>

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
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Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

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c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

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