7.71 grams sulfur (1 mole S/32.07 grams)
= 0.2404 moles of sulfur
Here are 0,2 moles sulfur.
3.3 moles of K2S 3.3 moles of S-2 6.6 moles of K+1
300 g sulfur is equivalent to 9,36 moles.
Oxidation-reduction reaction:Ag^+(aq) + Al(s) ===> Ag(s) + Al^3+ or looked at another way... 3AgNO3(aq) + Al(s) ===> Al(NO3)3(aq) + 3Ag(s) moles AgNO3 present = 92.8 g x 1 mole/170 g =0.546 moles moles Al present = 1.34 g x 1 mole/26.9 g = 0.0498 moles Al is limiting based on mole ratio of 3 AgNO3 : 1 Al moles Ag(s) produced = 0.0498 moles Al x 3 moles Ag/mole Al = 0.1494 moles Ag mass of Ag = 0.1494 moles Ag x 108 g/mole = 16.1 g Ag formed
Hydrogen reacts with sulfur in the proportions dictated by this equation:H2 + S --> H2S. The reactants are in a 1:1 proportion with each other, so 1.72 moles of hydrogen will react with 1.72 moles of sulfur.
1 g of ammonia (NH3) is equal to 0,059 mol.
Here are 0,2 moles sulfur.
3.3 moles of K2S 3.3 moles of S-2 6.6 moles of K+1
6,5 grams of sulfur S are equal to 0,203 moles.
6,35 moles of S contain 38,24059444195.10e23 sulfur atoms.
300 g sulfur is equivalent to 9,36 moles.
Oxidation-reduction reaction:Ag^+(aq) + Al(s) ===> Ag(s) + Al^3+ or looked at another way... 3AgNO3(aq) + Al(s) ===> Al(NO3)3(aq) + 3Ag(s) moles AgNO3 present = 92.8 g x 1 mole/170 g =0.546 moles moles Al present = 1.34 g x 1 mole/26.9 g = 0.0498 moles Al is limiting based on mole ratio of 3 AgNO3 : 1 Al moles Ag(s) produced = 0.0498 moles Al x 3 moles Ag/mole Al = 0.1494 moles Ag mass of Ag = 0.1494 moles Ag x 108 g/mole = 16.1 g Ag formed
Hydrogen reacts with sulfur in the proportions dictated by this equation:H2 + S --> H2S. The reactants are in a 1:1 proportion with each other, so 1.72 moles of hydrogen will react with 1.72 moles of sulfur.
Cu(s) + 2AgNO3(aq) ---> Cu(NO3)2(aq) + 2Ag(s) So you need half as many moles of Cu. Thus 5.8/2 = 2.9 moles of Cu are needed.
AgNO3 + NaCl ===> AgCl(s) + NaNO37 moles silver nitrate will produce 7 moles of silver chloride provided there is sufficient (at least 7 moles) of sodium chloride.
0,092 moles CaO
Assuming the reaction is S + O2 --> SO2, this equation is balanced as written, with everything in a 1:1 molar ratio. So, 67.1 moles of product would require 67.1 moles of O2 reactant.