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Question

The reaction between aluminium carbide and water takes place according to the following equation:
 Al4C3+12H2O3CH4+4Al(OH)3 
Calculate the volume fo methane measured at STP released from 14.4 g of aluminum carbide by excess fo water.
 


Solution

According to balance equation 
 Al4C3+12H2O3CH4+4Al(OH)3 
Molar mass of aluminium carbide is 144 g m o l to the power of negative 1 end exponent
Number of moles of Aluminium carbide =14.4g144g=0.1mol
​1 mol of aluminium carbide produce 3 moles of methane at STP.
1 mole = 22.4 L at STP
Therefore 3 mole = 3 ×  22.4 L  = 67.2 L of methane at STP​
Or 1 mole of Aluminium carbide produce 67.2 L of methane at STP
therefore 0.1 mol of Aluminium carbide will produce = fraction numerator 67.2 L over denominator 1 space m o l end fraction cross times 0.1 space m o l
                                                                                   = 6.72 L of methane


Chemistry
Concise Chemistry
Standard X

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