Given :
Compound A reacts with Compound B to form only one product, Compound C.
The usual percent yield of C in this reaction is 40%.
10.0 g of A are reacted with excess Compound B, and 6.4 g of Compound C
To Find :
The theoretical yield of C.
Solution :
We know, % yield is given by :
Putting given values , we get :
Therefore, theoretical yield of C is 16 g.
Hence, this is the required solution.
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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
25/2 and 96/X
CROSS MULTIPLY.
2x=2,400.
divide by 2.
x=1,200.
you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right
