Question

fill in the missing words
write a balanced equation for the reaction of metallic iron and hydrochloric acid hcl that produces iron(ii) chloride and hydrogen gas.
note to students: for molar mass values, use periodic table from the textbook. all answers must be reported with the correct number of significant figures.
blank grams of iron(ii) chloride are expected to be produced if 3.72 g of metallic iron has reacted completely.

Explanation:

Step1: Write the balanced chemical equation

$Fe + 2HCl=FeCl_{2}+H_{2}\uparrow$

Step2: Calculate the molar mass of Fe and $FeCl_{2}$

The molar mass of $Fe$ ($M_{Fe}$) from the periodic - table is approximately $55.85\ g/mol$, and the molar mass of $FeCl_{2}$ ($M_{FeCl_{2}}$) is $M_{Fe}+2\times M_{Cl}=55.85 + 2\times35.45=126.75\ g/mol$.

Step3: Calculate the number of moles of Fe

The number of moles of $Fe$ ($n_{Fe}$) is calculated using the formula $n=\frac{m}{M}$, where $m = 3.72\ g$ and $M = 55.85\ g/mol$. So, $n_{Fe}=\frac{3.72\ g}{55.85\ g/mol}\approx0.0666\ mol$.

Step4: Determine the mole - ratio between Fe and $FeCl_{2}$

From the balanced equation, the mole - ratio of $Fe$ to $FeCl_{2}$ is $1:1$. So, the number of moles of $FeCl_{2}$ ($n_{FeCl_{2}}$) produced is equal to the number of moles of $Fe$ reacted, $n_{FeCl_{2}} = 0.0666\ mol$.

Step5: Calculate the mass of $FeCl_{2}$ produced

Using the formula $m = n\times M$, where $n = n_{FeCl_{2}}=0.0666\ mol$ and $M = M_{FeCl_{2}} = 126.75\ g/mol$. So, $m_{FeCl_{2}}=0.0666\ mol\times126.75\ g/mol\approx8.44\ g$.

Answer:

$8.44$