How many moles of water are needed to react with 100. g of al4c3

The moles of water are needed to react with 100. g of al4c3 - 8.34 moles.

The reaction can be expressed as -

Al₄C₃ + 12 H₂O → 4 Al(OH)₃ + 3 CH₄

By the equation, it can be said that 1 equivalent of Al₄C₃ reacts with the equivalent of 12 H₂O to generate 4 Al(OH)₃ and 3 CH₄.

the moles = mass + molar mass

then for moles of Al₄C₃ = 100 + 143.959

= 0.6946 moles.

or, 216/143.959

for 100 gram = 21600/143.959

= 150. 0416

moles of H₂O required =

= 150.0416/18

= 8.43 moles.

Thus, the moles of water are needed to react with 100. g of al4c3 - 8.34 moles.

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8.4 moles

Explanation:

Al4C3 + 12 H2O ---> 4 Al(OH)3 + 3 CH4

100g of Al4C3 = 100/144 = 0.7 moles

from the reaction above,

1 mole of Al4C3 = 12 moles of water

0.7 moles of Al4C3 = x moles of water

x = 12×0.7/1 = 8.4 moles

Therefore 8.4 moles of water is needed to react with 100. g of al4c3

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m=9

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