Zinc reacts with aqueous sulfuric acid to form hydrogen gas: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) In an experiment, 201 mL of wet H2 is collected over water at 27 °C and a barometric pressure of 732torr. How many grams of Zn have been consumed? The vapor pressure of water at 27 °C is 26.74 torr.

0.4951 grams of Zn have been consumed.

Explanation:

Pressure at which gas is collected over water = p = 732 Torr

Vapor pressure of the water = p' = 26.74 Torr

Pressure of hydrogen gas = P

p = P + p'

P = p - p' = 732 Torr - 26.74 Torr = 705.26Torr

P =

1 atm = 760 Torr

Volume of the hydrogen gas = V = 201 mL = 0.201 L

1 mL= 0.001 L

Moles of hydrogen gas = n

Temperature at which hydrogen gas collected = T = 27°C = 27+273 K = 300 K

Using an ideal gas equation:

n = 0.007573 mol

According to recation , 1 mole of hydrgen gas was obtained from 1 mole of zinc,then 0.007573 moles of hydrgen gas will be obtained from;

of zinc

Mass of 0.007573 moles of zinc:

0.007573 mol × 65.38 g/mol = 0.4951 g

0.4951 grams of Zn have been consumed.

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answer; i believe the correct answer is (solid gold);