Question

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hydrochloric acid can dissolve solid iron according to the following reaction.
fe(s) + 2hcl(aq) → fecl₂(aq) + h₂(g)
part a
what minimum mass of hcl in grams would you need to dissolve a 2.6 g iron bar on a padlock?
express your answer using two significant figures.
part b
how much h₂ would be produced by the complete reaction of the iron bar?
express your answer using two significant figures.

Explanation:

Part A

Step1: Calculate the molar mass of Fe and HCl

The molar mass of Fe ($M_{Fe}$) is $55.85\ g/mol$, and the molar mass of HCl ($M_{HCl}$) is $1 + 35.5=36.5\ g/mol$.
From the balanced chemical equation $Fe(s)+2HCl(aq)\to FeCl_2(aq)+H_2(g)$, the mole ratio of Fe to HCl is $1:2$.

Step2: Calculate the moles of Fe

Given the mass of Fe ($m_{Fe} = 2.8\ g$), the moles of Fe ($n_{Fe}$) is calculated by $n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{2.8\ g}{55.85\ g/mol}\approx0.0501\ mol$.

Step3: Calculate the moles of HCl

Since the mole ratio of Fe to HCl is $1:2$, the moles of HCl ($n_{HCl}$) is $n_{HCl}=2n_{Fe}=2\times0.0501\ mol = 0.1002\ mol$.

Step4: Calculate the mass of HCl

Using the formula $m = nM$, the mass of HCl ($m_{HCl}$) is $m_{HCl}=n_{HCl}\times M_{HCl}=0.1002\ mol\times36.5\ g/mol\approx3.7\ g$.

Part B

Step1: Use the mole - ratio from the balanced equation

From the balanced equation $Fe(s)+2HCl(aq)\to FeCl_2(aq)+H_2(g)$, the mole ratio of Fe to $H_2$ is $1:1$.

Step2: Calculate the moles of Fe

We already know from Part A (using $m_{Fe} = 2.8\ g$ and $M_{Fe}=55.85\ g/mol$) that $n_{Fe}=\frac{2.8\ g}{55.85\ g/mol}\approx0.0501\ mol$.

Step3: Calculate the moles of $H_2$

Since the mole ratio of Fe to $H_2$ is $1:1$, $n_{H_2}=n_{Fe}\approx0.0501\ mol$.

Step4: Calculate the mass of $H_2$

The molar mass of $H_2$ ($M_{H_2}$) is $2\ g/mol$. Using $m = nM$, $m_{H_2}=n_{H_2}\times M_{H_2}=0.0501\ mol\times2\ g/mol = 0.10\ g$.

Answer:

Part A: $3.7\ g$
Part B: $0.10\ g$