student carried out this experiment using 0.0785 g of pure magnesium ribbon. At

Question

student carried out this experiment using 0.0785 g of pure magnesium ribbon. At the end of the reaction student also meas be 10.0 cm. (The vapor pressure of water at 220-19.8 tom:R=008206 Lam /mol K) to Calculate: (a) the pressure due to liquid column above water level in the beaker using the formula: P. sh x 0.735 tor/em, where h is height of liquid column in cm: (b) the partial pressure of H2 gas collected; (c) the volume of H2 gas collected at STP, and (d) the molar volume of H gas at STP 80.0 mL of H2 was collected at 22 and the barometric pressure was 755.0 torr. The ured the height (h) of liquid column in eudiometer (above the level in beaker) to

Explanation / Answer

For the given reaction,

Mg (s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g)

height of column = 10 cm

T = 22 oC + 273 = 295 K

Volume of H2 collected = 80 ml = 0.08 L

STP conditions, T = 273.15 K, P = 1 atm

a) Pressure due to liquid column (Pl) = h x 0.735 torr/cm

                                                           = 10 cm x 0.735 = 7.35 torr/cm

b) Partial pressure of H2 gas collected = 755 - 19.8 = 735.2 torr

c) moles of Mg = 0.0785 g/24.3 g/mol = 0.00323 mol

So, moles of H2 produced = 0.00323 mol

at STP conditions,

Volume of H2 collected at STP,

P1.V1/T1 = P2.V2/T2

with, P1, V1 and T1 be STP conditions

Thus,

Volume of H2 gas at STP (V1) = (735.2/760)atm(0.080)L(273.15K)/(1 atm)(295 K)

                                                  = 0.072 L

d) molar volume of H2 gas at STP = 0.072 L/0.00323 mol = 22.30 L

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