Question

When 20.4g of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride

produced? Explain (use calculations to prove your answer.
.ves or no
-with amounts shown)

Answer 1

Step-by-step explanation:

The answer: True

I explained the answer to your question in the submitted photo

Answer 2

Answer: True

Step-by-step explanation:

`\mathrm{The\ balanced \ chemical \ equation\ is\ :}\\$2 \mathrm{Na}{(s)}+\mathrm{Cl}_(2)(g) \rightarrow 2 \mathrm{NaCl}(s)`

For Na

`\\$Given mass $=20.4 \mathrm{~g}$ \ \ Molecular mass $=23 \mathrm{~g}\ \ $Number of moles $=\frac{\text { Given mass }}{\text { Molecular mass }}$$$\begin{aligned}&=(20.4)/(23) \\&=0.887 \text { moles }\end{aligned}$$Stoichiometric ratio of $\mathrm{Na}$ is same as that of $\mathrm{NaCl}$ i.e., $1: 1$.$\therefore$ Moles of $\mathrm{NaCl}=0.887$ motes`

For NaCl

`$$\\\begin{aligned}\text { Molecular mass } &=23+35.5 \\&=58.5 \mathrm{~g}\end{aligned}$$`

Therefore, it is true that when 20.4g of sodium metal are mixed with chlorine gas, 52.0 g of sodium chloride is produced