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small quantities of hydrogen gas can be prepared in the labo- ratory by the addition of aqueous hydrochloric acid to metallic zinc. zn1s2 1 2hcl1aq2 h zncl21aq2 1 h21g2 typically, the hydrogen gas is bubbled through water for col- lection and becomes saturated with water vapor. suppose 240. ml of hydrogen gas is collected at 30.8c and has a total pressure of 1.032 atm by this process. what is the partial pres- sure of hydrogen gas in the sample? how many grams of zinc must have reacted to produce this quantity of hydrogen? (the vapor pressure of water is 32 torr at 308c.)

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Final answer:

The partial pressure of hydrogen gas in the sample is 0.99 atm. The mass of zinc that reacted to produce this quantity of hydrogen gas is 0.784 g.

Step-by-step explanation:

The partial pressure of hydrogen gas in the sample can be determined by subtracting the vapor pressure of water from the total pressure. In this case, the total pressure is given as 1.032 atm, and the vapor pressure of water at 30.8°C is 32 torr. To convert the vapor pressure to atm, we divide by 760 (since 1 atm = 760 torr).

Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water

= 1.032 atm - 32 torr/760 torr/atm

= 1.032 atm - 0.042 atm

= 0.99 atm

To calculate the amount of zinc that reacted to produce this quantity of hydrogen gas, we can use stoichiometry. The balanced chemical equation for the reaction is:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

From the equation, we can see that 1 mole of zinc reacts to produce 1 mole of hydrogen gas. So, to find the amount of zinc, we need to calculate the number of moles of hydrogen gas using the ideal gas law and then use the mole ratio from the balanced equation.

First, let's calculate the number of moles of hydrogen gas:

n = PV/RT

= (0.99 atm)(0.24 L)/(0.0821 atm·L/mol·K)(30.8°C + 273.15 K)

= 0.012 moles of H2 gas

Since the mole ratio from the balanced equation is 1:1, the same amount of zinc must have reacted. Now, we can calculate the mass of zinc using the molar mass of zinc:

Mass = moles × molar mass

= 0.012 moles × 65.38 g/mol

= 0.784 g

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