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soldi70 [24.7K]
2 years ago
5

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:

Chemistry
1 answer:
Lerok [7]2 years ago
7 0

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

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How many molecules are in 3 moles of potassium bromide (KBr)
sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

KBr

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So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

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+

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≈

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−

So, if one mole of potassium bromide has a mas of

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m it follows that three moles will have a mass of

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moles KBr

⋅

molar mass of KBr



119 g

1

mole KBr

=

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You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

360 g

a

a

∣

∣

−−−−−−−−−

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