Asked by Ally
Help please:
A 0.199-g sample of unknown metal (X) reacts with hydrochloric acid to produce 88.5 mL of "wet" gas at 20 °C and 754 mm Hg. What is the unknown metal (X)? (The vapor pressure of water at 20 °C is 18 mm Hg.)
X(s) + 2 HCl(aq) → XCl2(aq) + H2(g)
(Would Mg be correct?)
A 0.199-g sample of unknown metal (X) reacts with hydrochloric acid to produce 88.5 mL of "wet" gas at 20 °C and 754 mm Hg. What is the unknown metal (X)? (The vapor pressure of water at 20 °C is 18 mm Hg.)
X(s) + 2 HCl(aq) → XCl2(aq) + H2(g)
(Would Mg be correct?)
Answers
Answered by
DrBob222
I don't think so.
Pdry gas = 754mm - 18 mm = 736 mm and
P gas in atm is 736/760 = ?
Then PV = nRT or
n = PV/RT = 736*0.0885/(760*0.08206*293) = ?
Then mols = g/atomic mass or
atomic mass = g/mols. Atomic mass Mg is about 24 and I obtained more than twice that.
Pdry gas = 754mm - 18 mm = 736 mm and
P gas in atm is 736/760 = ?
Then PV = nRT or
n = PV/RT = 736*0.0885/(760*0.08206*293) = ?
Then mols = g/atomic mass or
atomic mass = g/mols. Atomic mass Mg is about 24 and I obtained more than twice that.
Answered by
irishsooner
Would it be possible for someone to explain how you figure out which element is the answer? I'm clueless on this problem myself.
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