What mass in grams of AgCl is produced when 4.22g of AgNO3 react with 7.73g of AlCl3?
3AgNO3 (aq) + AlCl3 (aq) -
Al(NO3)3 (aq) + 3AgCl (s)

Question

What mass in grams of AgCl is produced when 4.22g of AgNO3 react with 7.73g of AlCl3?
3AgNO3 (aq) + AlCl3 (aq) -> 
Al(NO3)3 (aq) + 3AgCl (s)

ipwnbunnies · Answer

You probably have to first find the limiting reactant. Use the limiting reactant's amount of moles and use stoichiometry to figure out how much AgCl you can get from that. Then multiply the amount of AgCl by its molecular molar mass.

ipwnbunnies · Answer

Actually, when analyzing the limiting reactant, you'll already be solving for the least amount of AgCl, because that reactant will be the limiting reactant.

anonymous · Answer

I got 3.56g of AgCl

ipwnbunnies · Answer

Err, I'm not sure, I don't have a Table or a calc right now. Can you write out your process, or nah? lol

anonymous · Answer

Well I got .0248 mol of AgNO3 and .0580 mol AlCl3 meaning my limiting is AgNO3 so I multiply .0248 times the mass of AgCl 143.32g (.0248 * 143.32) = 3.56

ipwnbunnies · Answer

Timeout. Your limiting reactant will be the reactant that PRODUCES the LEAST amount of PRODUCT. So you have to use stoichiometry on both reactants first to figure out who makes the least amount of AgCl.

ipwnbunnies · Answer

Next, you need to use that amount of AgCl and multiply it by the mass of AgCl. You multiplied the amount of AgNO3 times the molecular mass of AgCl, which is useless.