# Question 9541d

Mar 12, 2015

Your reaction needed $\text{4.1 g}$ of iron.

$2 F {e}_{\left(s\right)} + 6 H C {l}_{\left(a q\right)} \to 2 F e C {l}_{3 \left(a q\right)} + 3 {H}_{2 \left(g\right)}$

Keep an eye out for the mole ratios that exist between iron and hydrochloric acid on one hand, and between hydrochloric acid and hydrogen gas on the other.

To solve this problem you have to work backwards from the amount of hydrogen gas produced, to the amount of iron required by using the number of moles of hydrochloric acid.

Since you know its concentration and volume, you can determine how many moles of $H C l$ were used in the reaction

$C = \frac{n}{V} \implies {n}_{\text{HCl}} = C \cdot V$
n_("HCl") = 500*10^(-3)"L" * "0.50 M" = "0.250 moles HCl"

In order to see whether or not all the moles of hydrochloric acid reacted, you must look at how many moles of hydrogen gas were produced. SInce you're at STP, use the fact that 1 mole of any ideal gas occupies 22.7 L under trese conditions - this is known as the molar volume of a gas at STP (100 kPa, 273.15 K)

n = V/V_("molar") = "2.50 L"/"22.7 L" = "0.1101 moles" ${H}_{2}$

The $\text{2:1}$ mole ratio that exists between hydrochloric acid and hydrogen gas can be used to see exactly how many moles of $H C l$ were consumed to produce this much hydrogen gas

$\text{0.1116 moles hydrogen gas" * "2 moles HCl"/"1 mole hydrogen gas" = "0.2202 moles HCl consumed}$

This means that not all the hydrochloric acid reacted, which implies that iron is a limiting reagent, i.e. it dictated how much of the hydrochloric acid actually reacted.

Since you know how much $H C l$ reacted, you can use the $\text{3:1}$ mole ratio between it and iron to see how many moles of iron were used

$\text{0.2202 moles HCl" * "1 mole Fe"/"3 moles HCl" = "0.0734 moles Fe}$

Use iron's molar mass to calculate the mass needed

$\text{0.0734 moles" * "55.845 g"/"1 mole" = "4.099 g Fe}$

If you round this to two sig figs, the number of sig figs in 0.50 M. the answer will be

m_("iron") = "4.1 g"#