Answer:
[tex]\boxed {\boxed {\sf About \ 16 \ grams \ of \ silver}}[/tex]
Explanation:
We are given the reaction:
[tex]Cu_{s}+AgNO_{3(aq)} \rightarrow CuNO_{3(aq)}+Ag_{s}[/tex]
We know that there are 25 grams of silver nitrate, or AgNO₃. First, we must find the molar mass of silver nitrate.
Silver Nitrate (AgNO₃)
Identify the molar masses of each element in silver nitrate using the Peirodic Table.
- Silver (Ag): 107.868 g/mol
- Nitrogen (N) : 14.007 g/mol
- Oxygen (O): 15.999 g/mol
Next, calculate the molar mass. There is a subscript of 3 after the O, so we must multiply oxygen's molar mass by 3.
- O₃= (15.999 g/mol) *3=49.997 g/mol
AgNO₃= (107.868 g/mol) + (14.007 g/mol)+(49.997 g/mol)=169.872 g/mol
Silver
Next, use stoichometry to find the mass of the silver.
- Convert grams of silver nitrate to moles.
[tex]25 \ g \ AgNO_3*\frac{1 \ mol \ AgNO_3}{169.872 \ g \ AgNO_3}\\[/tex]
[tex]\frac{25 \ mol \ AgNO_3}{169.872}= 0.1471696336 \ mol \ AgNO_3[/tex]
In this reaction, 1 mole of silver nitrate yields 1 mole of silver.
2. Convert moles of silver nitrate to moles of silver.
[tex]0.1471696336 \ mol \ AgNO_3*\frac{1 \ mol \ Ag}{1 \ mol \ AgNO_3}=0.1471696336 \ mol \ Ag[/tex]
We know that silver's molar mass is 107.868 grams per mole.
3. Convert moles of silver to grams of silver.
[tex]0.1471696336 \ mol \ Ag*\frac{107.868 \ g\ Ag}{1 \ mol \ Ag }[/tex]
[tex]0.1471696336 *{107.868 \ g\ Ag}=15.87489404 \ g\ Ag[/tex]
4. Round
The original measurement given had 2 siginficant figures (25= 2 and 5). Therefore, we must round to 2 sig figs or the nearest whole number for this problem.
The 8 in the tenth place tells us to round up to the nearest whole number.
[tex]15.87489404 \ g \ Ag \approx 16 \ g\ Ag[/tex]
About 16 grams of silver would be produced.
