Help me out, please?
A mass of 14.72g of (NH4)2 SO4 (molar mass = 132.06 g/mol) is dissolved in water. After the solution is heated, 30.19 g of Al2(SO4)3 * 18H2) (molar mass = 666.36 g/mol) is added. Calculate the theoretical yield of the resulting alum. Hint: this is a limiting reactant problem.

Question

Help me out, please?
A mass of 14.72g of (NH4)2 SO4 (molar mass = 132.06 g/mol) is dissolved in water. After the solution is heated, 30.19 g of Al2(SO4)3 * 18H2) (molar mass = 666.36 g/mol) is added. Calculate the theoretical yield of the resulting alum. Hint: this is a limiting reactant problem.

aaronq · Answer

1. write an equation and balance it

anonymous · Answer

Is this it?
(NH4)2 SO4 (aq) + Al2(SO4)3 (aq) + --> 2 NH4Al(SO4)2*12 H2O

thomaster · Answer

$$Al_{2}(SO_{4})_{3}⋅18H_{2}O+(NH_{4})2SO_{4}→2NH_{4}Al(SO_{4})_{2}⋅12H_{2}O)$$
$$Al_{2}(SO_{4})_{3}⋅18H_{2}O=666.4g/mol$$$$(NH4)_{2}SO_{4}=132.1g/mol$$$$2NH_{4}Al(SO_{4})_{2}⋅12H_{2}O=453.3g/mol$$

thomaster · Answer

and yes yours is correct :)

thomaster · Answer

Now you have to find the reaction ratio and determine which one is the limiting reactant.
Convert the mass of the limiting reactant to moles and use the ratio to find out how many moles of product is made. then use the MW of the product to calculate the mass of the product formed.