Q. 21 - An Introduction to General, Organic, and Biological Chemistry

Question

A metal (M) completely reacts with $34.8 mL$ of a $0.520 M$ $HCl$ solution to form $MC{l}_3$ and $H_2$.

a. Write the balanced chemical equation for the reaction of the metal and $HCl$ 

b. What volume, in milliliters, of $H_2$ produced at STP?

c. How many moles of metal M reacted?

d. If the metal has a mass of $0.420 g$, use your results from part c to determine the molar mass of the metal M.

e. What are the name and symbol of metal M in part d?

f.  Write the balanced chemical equation using the symbol of the metal from part e. 

Step-by-Step Solution

Answer

Part(a) The balanced chemical equation is  $2\mathrm{M}\left(\mathrm{s}\right) +6\mathrm{HCl}\left(\mathrm{aq}\right)\to 2{\mathrm{MCl}}_3\left(\mathrm{aq}\right) + 3{\mathrm{H}}_2\left(\mathrm{g}\right)$
Part(b) $203 mL H_2$ is produced at STP.
Part(c) $6.03\times {10}^{-3}$ moles of metal reacted.

Part(d)  Molar mass of metal is $69.7 g/mole$ 

Part(e) The name of metal is Gallium (Ga)
Part(f) The balanced chemical equation is $2Ga\left(s\right) + 6HCl\left(aq\right) \stackrel{}{\to } 2GaC{l}_3\left(aq\right) + 3H_2\left(g\right)$

Part(a) Step 1 : Given information

We are given that a metal reacts with with $HCl$.

Part(a) Step 2 : Simplify

As we know, the metal is reacting with $HCl$, therefore, the balanced chemical reaction is $2\mathrm{M}\left(\mathrm{s}\right) +6\mathrm{HCl}\left(\mathrm{aq}\right)\to 2{\mathrm{MCl}}_3\left(\mathrm{aq}\right) + 3{\mathrm{H}}_2\left(\mathrm{g}\right)$

Part(b) Step 1 : Given information

We are given that a metal reacts with with $HCl$.

Part(b) Step 2 : Simplify

From above equation We can say millimoles of $HCl$ is two times millimoles of $H_2$ 

And we also know volume at STP is $22.4 L$

Therefore, volume of $H_2$ produced at STP is $203 mL$

Part(c) Step 1 : Given information

We are given that a metal reacts with with $HCl$

Part(c) Step 2 : Simplify

From the balanced equation, We can say that millimoles of metal is equal to three times millimoles of $HCl$.

millimoles of $HCl$ is product of molarity and volume Therefore,

millimoles of $HCl$ is $M_1{V}_1=34.8\times 0.520=18.096$

So, millimoles of metal is $\frac{1}{3}{M}_1{V}_1=\frac{1}{3}\times 18.096=6.03$

Hence, Moles of metal is $6.03\times {10}^{-3}$

Part(d) Step 1 : Given information

We are given that a metal reacts with with $HCl$.

Part(d) Step 2 : Simplify

As given mass of metal is $0.420 g$

Moles of metal is $6.03\times {10}^{-3}$

Molar mass is $\frac{given mass}{number of moles}=\frac{0.420}{6.03\times {10}^{-3}}=69.7 g/mole$

Part(e) Step 1 : Given information

We are given that a metal reacts with with $HCl$.

Part(e) Step 2 : Simplify

From above part , we can say the name of metal is Gallium and its symbol is $Ga$.

Part(f) Step 1 : Given information

We are given that a metal reacts with with $HCl$.

Part(f) Step 2 : Simplify

As we know from part(e) that the metal is Gallium.

Therefore, by replacing metal in first equation by gallium.

So, the balanced equation is : $2Ga\left(s\right) + 6HCl\left(aq\right) \stackrel{}{\to } 2GaC{l}_3\left(aq\right) + 3H_2\left(g\right)$