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What volume of hydrogen gas would be produced if 2.00 x 10^22 atoms of magnesium were...
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First, we have to generate the reaction equation. For the reaction of Magnesium metal with HCl solution, MgCl2 (magnesium chloride) and H2 gas is produced. The balanced chemical equation can be written as:
Mg (s) + 2HCl (aq) ---> MgCl2 (aq) + H2 (gas)
To solve for the amount of H2 gas produced, we should follow these steps:
Atoms Mg -> moles Mg -> moles H2 -> volume of H2 (at STP; 273.15K and 1 atm)
`2.00x10^(22) at oms Mg * ( (1 mol e Mg)/(6.022x10^(23) )) * ( (1 mol e H2)/(1 mol e Mg) ) = 0.033211 mol es H2`
To get the volume of H2 gas at STP, we will employ the Ideal gas equation.
`PV = nRT`
`V = (nRT)/(P)`
`V = (0.033212 mol es * 0.08206 * 273.15)/(1atm)`
`V = 0.744 L H2 gas`
Posted by jerichorayel on December 28, 2012 at 8:42 AM (Answer #2)
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