Question 6. a. What mass of water in grams will form when 2.70 mL of oxygen gas

Question

Question 6.

a. What mass of water in grams will form when 2.70 mL of oxygen gas is produced ?

b.What volume of oxygen gas in liters will be produced from 0.405 g of hydrogen peroxide?

Testing for oxygen (I) obtain hydrogen peroxide solution and manganese (r) oxide tube maio, (2) Carefully p cm of hydrogen peroxide into a small test reaction. peroride (3) Add a small about 2 to the hydrogen the hydrogeu amount of manganese (r) oxide is only added to the rate of the reaction decomposes into water and oxygen gas The oxide in table below the of (4) Watch the solution for approximately one mmnute and record your the into flame indicates presence (3) Bring a glowing wood splint to the opening of the test tube. A glowing splint that bursts oxygen gas Record your observations in the table below. (5) Discard the contents of the rest tube into the s

Explanation / Answer

2H2O2 -------> 2H2O + O2

a]

Given 2.7 ml of O2 is produces

PV = nRT [V in L and P in atm]

At STP ; P =1atm , V = 2.7 ml ; T = 298 K and R = 0.0821

n = PV /RT = 1.1035*10^-4 moles

1.1035*10^-4 moles of O2 is produce

From the reaction

When 1 moles of O2 is produces 2 moles of H2O is also produced

so When 1.1035*10^-4 moles of O2 is produced 2*1.1035*10^-4 moles of H2O is also produced

Mass of H2O produceed = Moles *MW = 2*1.1035*10^-4 *18 = 0.00397 gms i

b]

Moles of H2O2 = 0.405 / 34 = 0.0119

2 moles of H2O2 produces 1 mole of O2

0.0119 moles of H2O2 produces 0.0119/2 =5.955*10^-3 moles of O2

Using

PV = nRT

n = 5.955*10^-3 ; P =1 atm , V = ?

T = 298 K

V = 0.1457 L of O2 is produced

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