3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷24,4g/mol=0,5mol-limiting reagentn
(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.
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
7 Valence lectrons I am pretty sure
The theoretical yield is 0.0110 g H₂.
<em>Moles of H₂O</em> = 0.0987 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.005 477 mol H₂O
<em>Moles of H₂</em> = 0.005 477 mol H₂O × (2 mol H₂/2 mol H₂O) = 0.005 477 mol H₂
<em>Theoretical yield</em> of H₂ = 0.005 477 mol H₂ × (2.016 g H₂/1 mol H₂)
= 0.0110 g H₂
