______ How many liters of dry hydrogen gas at STP would be liberated by 0.635 gr

Question

______ How many liters of dry hydrogen gas at STP would be liberated by 0.635 grams of calcium metal? ______ How many liters of dry hydrogen gas at STP would be liberated by 0.553 grams of potassium metal? __ A sample of metal weighed 0.688 grams. If 108 mL of dry hydrogen gas at STP is evolved, what is the equivalent weight of this metal? _____ A student observed that 1.019 grams of a metal produced 106 mL of wet hydrogen measured with an eudiometer at 27.0 degree of Celcius and barometric pressure of 757 torr. Calculate the equivalent weight of the metal as measured in this experiment. The vapor pressure of water at 27.0 degree of Celcius is 26.7 torr.

Explanation / Answer

Ans 1 .

Ca + 2HCl = CaCl2 + H2

1 mole of calcium releases 1 mole of hydrogen gas at STP.

1 mole of gas occupies 22.4 litres of volume .

0.635 g of Calcium = 0.635 / 40.078 = 0.0158 moles

So number of moles of hydrogen gas = 0.0158 moles

Volume occupied by 0.0158 moles of H2 = 0.0158 x 22.4 = 0.355 L

or 355 mL

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