Question

If 5.0 moles of an ideal gas occupies a volume of 9.3 L, what volume will 15.0 moles of the gas occupy?

A mixture of oxygen, hydrogen, carbon dioxide and methane gases has a total pressure of 1.3 atm. If the partial pressure of hydrogen is 2.4 psi, the partial pressure of carbon dioxide is 110 torr and the partial pressure of methane is 0.22 atm, how many mmHg does the oxygen exert?

For the reaction below, how many mL of gas can be produced from 50 g of iron at STP?

2Fe(s) +6HCl(aq) →2FeCl3(aq) +3H2(g)

Answer 1

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5 mole gas occupy volume 6.3 L then 15 mole gas occupy volume = 15 6.3 / 5 = 18.9 L

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According to dalton's law total pressure of gas is equal to sum of partial pressure exerted by each individual gas

P_total = P₁ + P₂ + P₃

P_oxygen = P_total - P_CO2 - P_H2 - P_metane

2.4 psi = 0.16331 atm

110 torr = 0.144737 atm

substitute value

P_oxygen = 1.3 - 0.16331 - 0.144737 - 0.22 = 0.771953 atm

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molar mass iron = 55.85 gm/mole then 2 mole of iron = 55.85 2 = 111.7 gm

molar mass of H₂ = 2.01 gm/mole then 3 mole of H₂ = 2.01 3 = 6.03 gm

Accordng to reaction 2 mole of iron produce 3 mole of H₂ that mean 111.7 gm of iron produce 6.03 gm of hydrogen

then 50 gm of iron produce = 50 6.03 / 111.7 = 2.7 gm of H₂

molar mass of H₂ = 2.01 gm/mole then 2.7 gm of H₂ = 2.7 / 2.01 = 1.34 mole of H₂

at STP 1 mole of gas occupy volume 22.414 liter then 1.34 mole of gas occupy volume = 1.34 22.414 = 30.03475 liter = 30034.75 ml

H₂ occupy volume = 30034.75 ml