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Solid sodium reacts with liquid water to form hydrogen gas according to the equation 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) What is the pressure of hydrogen gas in the 20.0 L headspace of a reactor vessel when 1.00 kg sodium is reacted with excess water at 50.0°C?

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mickymike92

Answer: 21850.42mmHg or 28.78atm

Explanation: From the equation of the reaction, 2 moles of sodium reacts with 2 moles of water to produce 2 moles of NaoH and 1 mole of hydrogen gas at s.t.p.

molar volume of gas at s.t.p. = 22.4dm^3,

standard temperature = 0°C or 273K,

standard pressure = 760mmHg or 1atm,

molar mass of Na = 23g;

volume of the container= 20L or 20dm^3

From the equation, 2 moles or 46g of Na produces i mole or 22.4 dm^3 of hydrogen.

Therefore, 1kg or 1000g of Na will produce 22.4 * 1000/(46)= 486dm^3 of hydrogen.

To find the pressure of the gas produced, the formula P1V1/T1 = P2V2/T2

P1= 760mmHg, T1= 273K, V1 (initial volume of gas)= 486dm^3, T2=50.0°C or 50+273K= 323K, V2( final volume of gas= volume of vessel)=20dm^3  P2=?

P1= P1V1*T2/(V2T1)= 760mmHg * 323/(273)= 21850.42mmHg or 1atm*21850.42/760= 28.78atm

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