23,748 views
2 votes
2 votes
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)

User Brian Donahue
by
3.1k points

2 Answers

13 votes
13 votes

Step-by-step explanation:

The answer: True

I explained the answer to your question in the submitted photo

When 20.4g of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride-example-1
User Roger Trullo
by
3.6k points
16 votes
16 votes

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


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


\begin{aligned}\\& $\therefore$ Mass of $\mathrm{NaCl}$ produced will be=\text { No.of moles } * \text { Molecular mass } \\&=0.887 * 58.5 \\&=51.9 \\&\simeq 52 \mathrm{~g} \quad\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

User Patrick Wright
by
3.1k points