198k views
5 votes
Please help me with stoichiometry

Please help me with stoichiometry-example-1

1 Answer

4 votes

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.

User Grenzbotin
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.