Answer:
2. 6.07 g
3. 38%
Step-by-step explanation:
2.
2AgNO3(aq) + Na2S(aq) --> Ag2S(s) + 2NaNO3(aq)
molecular mass of AgNO3 is 169.87 g/mol
so converting 7.79 g into mole : 7.79/169.87 = 0.049 mol
2 mol of AgNO3 will produce 1 mole of Ag2S
so 0.049 mole of AgNO3 will produce 0.049/2 = 0.0245 mol of Ag2S
since molecular mass of Ag2S is 247.8 g/mol
=> 0.0245 x 247.8 = 6.07 g
3.
molecular mass of NaBr is 102.894 g/mol
=> 3 gm will have: 3.0 / 102.849 = 0.029 mol
according to the reaction, 1 mole of NaBr will produce 1 mole of HgBr
so 0.029 mol of NaBr will produce 0.029 mol of HgBr
since molecular mass of HgBr is 360.41 g/mol
=> mass = 0.029 x 360.41 = 10.45 g
so actual mass/theoretical mass
4.000/10.45 = 0.382 or 38%
Please double check the calculations. Hope this help.