Question

3. Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water what mass of
reactant remains?
The unbalanced equation is:
AI2S3 + H2O -> Al(OH)3 + H2S​

Answer 1

Answer:

Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?

The unbalanced equation is: Al2S3 + H2O ----> Al(OH)3 + H2S

B) How many grams of PF5 can be formed from 9.46 g of PF3 and 9.42 g og XeF4 in the following reaction?

2PF3 + XeF4 ----> 2PF5 + Xe