Question

In: Chemistry

Part A: A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion...

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

Solutions

Expert Solution

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


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