Part A) Calculate the mass of water produced when 6.23 g of butane reacts with e

Question

Part A) Calculate the mass of water produced when 6.23 g of butane reacts with excess oxygen.

Part B) Calculate the mass of butane needed to produce 78.5 g of carbon dioxide.

Part C) A sample of sodium reacts completely with 0.355 kg of chlorine, forming 585 g of sodium chloride. What mass of sodium reacted?

Part D) When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of 92.2 g of oxygen, 18.6 g of oxygen remained unreacted. What mass of carbon dioxide was produced?

Explanation / Answer

Ans :

Part A )

The reaction is given as :

C4H10 + 7/2 O2 = 4CO2 + 5H2O

6.23 g of butane = 6.23 / 58.12 = 0.107 moles

0.107 moles of butane will produce 5 x 0.107 = 0.535 moles of water

Mass of water = 0.535 x 18.016

= 9.64 grams

Part B)

78.5 g CO2 = 78.5 / 44.01 = 1.78 moles

Number of moles of butane required = 1.78 / 4 = 0.445 moles

Mass of butane required = 0.445 x 58.12

= 25.86 grams

Part C :

The reaction is given as :

Na + 1/2 Cl2 = NaCl

Number of moles of NaCl = 585 / 58.44 = 10.01 moles

So number of moles of sodium reacted = 10.01 moles

mass of sodium reacted = 10.01 x 22.99

= 230.13 grams

Part D :

The reaction is given as :

C + O2 = CO2

mass of oxygen reacted = 92.2 - 18.6 = 73.6 grams

Number of moles of O2 = 73.6 / 31.99 = 2.30 moles

number of moles of C = 27.6 / 12.01 = 2.30 moles

So number of moles of CO2 formed will also be 2.30

Mass of CO2 formed = 2.30 x 44.01

= 101.223 grams

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