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Black_prince [1.1K]
3 years ago
6

This is my chemistry worksheet. It's a new topic my teacher will barely review with us. Help please???

Chemistry
1 answer:
TEA [102]3 years ago
6 0
So to balance an equation, you need to get the same amount of each type of element on either side of the --> . So you pretty much are given the subscripts in the equations and you need to add coefficients (just normal numbers) in front of any formula that needs it, keeping anything balance.
KCl_{3}  + O_{2}  -\ \textgreater \  KCl_{3}
turns into
2KCl_{3}+ 3O_{2} -> 2KCl_{3}

These coefficient numbers are the molar ratios, so 2 moles of KCl3 for every 3 moles of O2   so 1. 3:2

Then you can use these ratios of find out how many moles of one thing are needed if you are given the amount of another.
\frac{moles of element 1}{cofficient 1}  =  \frac{moles of element 2}{cofficient 2}
and use cross multiplication to solve for whatever you don't know


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A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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