A sampleof gas containing 63.4g of oxygen gas an initial volume of 7.51L.What is the final volume in L if 4.6moles of oxygen gas is added?
- hcbiochemLv 73 weeks ago
Initial moles O2 = 63.4 g / 32.0 g/mol = 1.98 mol O2
Final moles O2 = 6.58 mol
V1/n1 = V2/n2
7.51 L / 1.98 mol = V2 / 6.58 mol
V2 = 25 L
(Your answer is limited to 2 significant figures because of the added moles of O2)
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