How many grams of aloh3 will be produced from 0.734moles of ch4

 

Answer: 0.734 moles of CH4 will produce 114.50 grams of Al(OH)3.

To determine the number of grams of Al(OH)3 produced from 0.734 moles of CH4, we need to use the balanced chemical equation that relates the two compounds, and then use stoichiometry to convert from moles of CH4 to moles of Al(OH)3 and finally to grams of Al(OH)3.

The balanced chemical equation for the reaction between CH4 and Al(OH)3 is:

CH4 + 4Al(OH)3 → 2Al(OH)3 + CH4O

From the equation, we can see that for every 1 mole of CH4, 2 moles of Al(OH)3 are produced.

So, starting with 0.734 moles of CH4, we can calculate the number of moles of Al(OH)3 produced:

Moles of Al(OH)3 = 0.734 moles CH4 x (2 moles Al(OH)3 / 1 mole CH4)

= 1.468 moles Al(OH)3

Now that we know the number of moles of Al(OH)3 produced, we can calculate the mass of Al(OH)3 using its molar mass:

Molar mass of Al(OH)3 = (1 x atomic weight of Al) + (3 x atomic weight of O) + (3 x atomic weight of H)

= (1 x 26.98 g/mol) + (3 x 16.00 g/mol) + (3 x 1.01 g/mol)

= 78.00 g/mol

Mass of Al(OH)3 = Moles of Al(OH)3 x Molar mass of Al(OH)3

= 1.468 moles x 78.00 g/mol

= 114.50 g

Therefore, 0.734 moles of CH4 will produce 114.50 grams of Al(OH)3.