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3 years ago

How many grams of sodium metal are needed to make 29.3 grams of sodium chloride? given the reaction: 2na + cl2 → 2nacl?

Chemistry

2 answers:

Bess [88]3 years ago

5 0

2Na + Cl2 → 2NaCl
46 g. ➡️ 117 g
x. ➡️ 29.3 g
$x = \frac{46 \times 29.3}{117} \\ x = 11.5 \: g$

GaryK [48]3 years ago

5 0

Answer:

11.52 grams of sodium

Explanation:

The balanced equation is :

$2Na+Cl_{2}-->2NaCl$

As per balanced equation, two moles of sodium will give two moles of sodium chloride.

the molar mass of sodium chloride = atomic mass of Na + atomic mass of Cl

The molar mass of sodium chloride = 23 +35.5 = 58.5

thus for 58.5 grams of sodium chloride the mass of sodium required = 23 g

For one gram of sodium chloride the mass of sodium required = $\frac{23}{58.5}=0.393g$

Therefore for 29.3 grams of sodium chloride = 29.3x0.393 g of sodium = 11.52g

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$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

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3 years ago