112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below. 2 Al4C3(s) + 12 H2O(l) → 3 CH4(g) + 4 Al(OH)3(s)

If aluminum carbide is the limiting reactant, how many moles of the excess reactant are left over
a
37.3 mol
b
4.68 mol
c
7.33 mol
d
131.94 mol

4.999 moles of excess reactant will be left over.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

      .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

For the given chemical reaction:

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be