A student carried out this experiment using 0.0795 g of pure magnesium ribbon. A

Question

A student carried out this experiment using 0.0795 g of pure magnesium ribbon. At the end of the reaction 81.4 mL was collected at 22 degree C and the barometric was 755.0 torr. The be 5.0 also measured pressure the level in beaker) t (a) cm. the height (h) quid column in eudiometer (above the level in breaker)be 5.0 cm. (vapor pressure of water at 22 degree C = 19.8 torr R = 0.08206/mol K) (a) Calculate the partial pressure of H_2 gas collected.(b) Calculate the volume of H_2 gas collector under STP condition. (c) Calculate the molar volume of H_2 gas at STP from this reaction.

Explanation / Answer

From the given experiment

(a) Partial pressure of H2 = 755 - 19.8 = 735.2 torr = 0.97 atm

(b) Volume of H2 at STP = P1V1T2/P1

with,

P1, V1 and T1 be pressure, volume and temperature of H2 from this experiment

T2 and P2 are temperature and pressure of at STP

Volume of H2 at STP = 0.97 x 0.0814 x 273.15/295 x 1

                                   = 0.073 L

(c) Mg + 2HCl ---> MgCl2 + H2

moles of Mg = 0.0795/24.32 = 0.0033 mol

So,

molar volume of H2 at STP = 0.073/0.0033 = 22.15 L

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