The reaction is:
Al2S3 + 6AgNO3(aq) --> 3Ag2S(s) + 2Al(NO3)3(aq)
Find the formula mass (molar mass) of Al2S3.
Divide 300.23 g by the formula mass to get moles of Al2S3.
Moles of Ag2S = (__moles Al2S3)(3)
Moles of Al(NO3)3 = (__moles Al2S3)(2)
In a certain reaction, 300.23 grams of solid aluminum sulfide combine with unlimited aqueous silver nitrate. How many moles of aluminum sulfide react? How many moles of each product will be produced?
First of all, is the reaction a double replacement or not? I have so much trouble with chem! Please help!
2 answers
double replacement.
YOu always start with the balanced chem reaction:
Al2S3 + 6AgNO3 >>3Ag2S+2Al(NO3)3
and the silver sulfide precipates out of solution.
So for each mole of aluminum sulfide, you get three moles of silver sulfide, and two moles of aluminum nitrate.
So how many moles of aluminum sulfide are in 300.23 grams? You get three times that number of moles of Ag2S, and twice that number of moles of Aluminum nitrate.
YOu always start with the balanced chem reaction:
Al2S3 + 6AgNO3 >>3Ag2S+2Al(NO3)3
and the silver sulfide precipates out of solution.
So for each mole of aluminum sulfide, you get three moles of silver sulfide, and two moles of aluminum nitrate.
So how many moles of aluminum sulfide are in 300.23 grams? You get three times that number of moles of Ag2S, and twice that number of moles of Aluminum nitrate.