In: Chemistry
Part A:
A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O2 gives 27.1 g of a mixture of CO2 and SO2.
Find the mass of sulfur in the original mixture.
Part B:
Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).
Thanks
Part A:
The reactions are:
S + O2 ----> SO2
C + O2 ----> CO2
Now, let´s say X is the mass of sulfur:
9-X= mass of C
Let´s find moles of sulfur and carbon:
mol= mass/MW
MW S= 32g/mol
MW C= 12g/mol
mol S= X/32g/mol
mol C= 9-X/12g/mol
1 mol of S produce 1 mol SO2, and 1 mol C produce 1 mol CO2:
mass SO2= X/32g/mol x 1mol SO2/1 mol S x 64.1g SO2/mol = 2Xg SO2
mass CO2= 9-X/12g/mol x 1 mol CO2 / 1 mol C x 44g CO2/mol = 33-3.667Xg CO2
27.1g= 2Xg SO2 + 33-3.667Xg CO2
X= 3.54g= mass of sulfur
Part B:
MW= 107.87 g/mol
In 1 mol of atoms there are 6.02x1023 atoms so:
107.87g/mol x 1 atom/6.02x1023 atoms x 1mol= mass of 1 atom of Ag
mass of 1 atom of Ag= 1.79x10-22g