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Sagot :
Answer:
212 gm of CH2Cl2 are formed
Explanation:
4.6 CONCEPT ASSESSMENT Suppose that each of these reactions has a 90% yield. CH4(g) + Cl2(g) - CH2Cl(g) + HCl(g) CH3Cl(g) + Cl2(g) - CH2Cl20) + HCl(g) Starting with 50.0 g CH4 in the first reaction and an excess of Cl2(g), the number of grams of CH2Cl2 formed in the second reaction is (a) 50.0 X 0.81 x (85/16) (c) 50.0 X 0.90 X 0.90 (b) 50.0 X 0.90 (d) 50.0 X 0.90 X 0.90 < (16/50.5)(70.9/85)H
1st, rewrite the equations so you can see the problem easier
reaction (1) CH4 +Cl2--------->CH3CL + HCl
reaction (2) CH3Cl + Cl2------> CH2Cl2 + HCl
the product ib reaction (1) is CH3Cl and is formed with a 90% yield.
90% of the reactant methane iis converted to methyk chloride.
methane (CH4) has a molar mass of 12 + 4X1 =16
you start with 50.0 gm of CH4,
SO
50.0/16 =3.125 moles of CH4
90% of this become methyl choride...CH3Cl
SO
0.90 X 3.125 = 2.8 moles of CH3Cl are formed
NOTE 90% HAS ONLY 2 SIGNIFICANT FIGURES SO YOU ARE LIMITED TO 2 SIGNIFICANT FIGURES IN YOUR CALCULATION!
in reaction(2) 1 mole of methyl chloride becomes methyl dichloromethane yi
with again a 90% yield
0.90 X 2.8 = 2.5 moles of CH2Cl2 are formed
so
2.5 moles of CH2Cl2 are formed.
CH2Cl2 has a molar mass of 12 + 2X1 + 2X35.5 =
12 + 2 + 71 =85
1 mole weighs 85 gm
2.5 moles weigh 2.5 X 85 = 212 gm
X
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