Respuesta :
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
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.
