Question

Part A:

Answer 1

Answer:

A. 3.53 g of Sulphur.

B. 1.79 x 10^-22 g of Silver atom.

Explanation:

A.

Equation of the reaction

C + S + 2O2 --> CO2 + SO2

Molar mass of:

C = 12 g/mol

S = 32 g/mol

CO2 = 12 + (16*2)

= 44 g/mol

SO2 = 32 + (32*2)

= 64 g/mol

Mass of C + S = 9 g

Mass of CO2 + SO2 = 27.1 g

Therefore, mass of C = 9 - mass of S;

Mass of C * ratio of molar mass of CO2 to C + Mass of S * ratio of molar mass of SO2 to S,

(Moles of C*molar mass of CO2) + (Moles of S*molar mass of SO2) = 27.1

(9 - S) * (44/12) + S* (64/32)

(9 - S) * (3.67) + S * 2

33 - 3.67S + 2S = 27.1

-1.67S = - 5.9

Mass of S = 3.53 g

Mass of C = 9.0 - 3.53

= 12.53 g

The mass of sulfur is 3.53 g.

B.

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Number of moles of 1 silver atom using Avogadros constant, 6.022 x 10^23 molecules per moles

Moles = 1/ 6.023 x 10^23

= 1.66*10^-24 moles

Mass = number of moles * molar mass

Mass = 1.66 x 10 ^-24 * 107.87

Mass = 1.79 x 10^-22 g of Silver atom.

Answer 2

Answer:

Part A = The mass of sulfur is 6.228 grams

Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams

Explanation:

Part A

Step 1: Data given

A mixture of carbon and sulfur has a mass of 9.0 g

Mass of the product = 27.1 grams

X = mass carbon

Y = mass sulfur

x + y = 9.0  grams

x = 9.0 - y

x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6

(9 - y)*(44.01/12.01) + y(64.07/32.07)

(9-y)(3.664) + y(1.998)

32.976 - 3.664y + 1.998y = 22.6

-1.666y = -10.376

y = 6.228 = mass sulfur

x = 9.0 - 6.228 = 2.772 grams = mass C

The mass of sulfur is 6.228 grams

Part B

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Calculate moles of 1 silver atom

Moles = 1/ 6.022*10^23

Moles = 1.66*10^-24 moles

Mass = moles * molar mass

Mass = 1.66*10 ^-24 moles *107.87

Mass = 1.79 * 10^-22 grams

The mass of 1 silver atom is 1.79 * 10^-22 grams