How many liters of hydrogen gas are produced at STP from the decomposition of $70.0 g$ $"70.0 g"$ of hydrogen chloride?

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How many liters of hydrogen gas are produced at STP from the decomposition of $70.0 g$ $"70.0 g"$ of hydrogen chloride?

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Answer 1 · 2015-01-18T19:17:39

The answer is $21.5 L$ $"21.5 L"$ .

So, start with the balanced chemical equation for the decomposition of hydrochloric acid

$2 HCl \to H_{2} + {Cl}_{2}$ $2"HCl" -> "H"_2 + "Cl"_2$

Notice that you have a $2:1$ $"2:1"$ mole ratio between $HCl$ $"HCl"$ and ${Cl}_{2}$ $"Cl"_2$ , which means that every 2 moles of the former will produce 1 mole of the latter. The number of moles of $HCl$ $"HCl"$ you have is

$70.0 g \cdot \frac{1 mole}{36.5 g} = 1.92$ $"70.0 g" * ("1 mole")/("36.5 g") = 1.92$ $moles$ $"moles"$

As a result, the number of ${Cl}_{2}$ $"Cl"_2$ moles is

$n_{C l_{2}} = \frac{n_{H C l}}{2} = \frac{1.92}{2} = 0.96$ $n_(Cl_2) = n_(HCl)/2 = 1.92/2 = 0.96$ $moles$ $"moles"$

At STP, 1 mole of any ideal gas occupies 22.4 L, which means that the volume of ${Cl}_{2}$ $"Cl"_2$ produced will be

$n = \frac{V}{V_{molar}} \Rightarrow V = n \cdot V_{molar} = 0.96$ $n = V/V_("molar") => V = n * V_("molar") = 0.96$ $moles \cdot 22.4$ $"moles" * 22.4$ $\frac{L}{mole} = 21.5$ $L/("mole") = 21.5$ $L$ $"L"$

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How many liters of hydrogen gas are produced at STP from the decomposition of $70.0 g$ $"70.0 g"$ of hydrogen chloride?

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How many liters of hydrogen gas are produced at STP from the decomposition of 70.0 g"70.0 g" of hydrogen chloride?

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How many liters of hydrogen gas are produced at STP from the decomposition of $70.0 g$ $"70.0 g"$ of hydrogen chloride?