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Triss [41]
3 years ago
9

In a 71.4 g sample of nahco 3 3 ​ , how many grams of sodium are present?

Chemistry
1 answer:
sweet [91]3 years ago
4 0

There are 19.5 g Na in 71.4 g NaHCO₃

Calculate the <em>molecular mass of NaHCO₃</em>.

1 Na = 1 × 22.99 u = 22.99   u

1 H   = 1 × 1.008 u =    1.008 u

1 C   = 1 × 12.01  u =  12.01    u

3 O = 3 × 16.00 u =  <u>48.00  u </u>

               TOTAL =  84.008 u

So, there are 22.99 g of Na in 84.008 g NaHCO₃.

∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na

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Answer :

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Explanation :

The given chemical reaction is:

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Initial conc.      0.1550      0.173           0

At eqm.          (0.1550-x)  (0.173-x)         x

As we are given:

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The expression for equilibrium constant is:

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get:

255=\frac{(x)}{(0.1550-x)\times (0.173-x)}

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We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

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The equilibrium concentration of COCl₂ = x = 0.139 M

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