Question

In the reaction of aluminum hydroxide and caldium nitrate, how many grams of calcium hydroxide will be formed if 12.55 g of aluminum hydroxide are used with excess calcium nitrate?

Answer 1

17.76g

Step-by-step explanation:

We need to write a balanced chemical equation for the reaction:

2Al(OH)3 + 3Ca(NO3)2 ——> 3Ca(OH)2 + 2Al(NO3)3

In the reaction above, it can be seen that 2 moles of aluminum hydroxide yielded 3 moles of calcium hydroxide. This is the theoretical viewpoint.

Now we need to know what actually happened. We need to calculate the actual number of moles of aluminum hydroxide reacted l. We can get this by dividing the mass by the molar mass.

The molar mass of aluminum hydroxide is 27+ 3( 16+1)

= 27 + 51 = 78g/mol

The number of moles is thus: 12.55/78 = 0.16 moles

Now if 2 moles of aluminum hydroxide gave 3 moles of calcium hydroxide, 0.16moles will give : (0.16*3)/2 = 0.24moles

Now we can calculate the mass of calcium hydroxide formed. The mass of calcium hydroxide formed is the number of moles multiplied by the molar mass.

The molar mass of calcium hydroxide is; 40 + 2(17) = 74g/mol

The mass is thus =74 * 0.24 = 17.76g