Question

A student added 7.40 X 10-2g of magnesium ribbon to 15 cm3 of 2.00 mol dm-3
hydrochloric acid. The hydrogen gas produced was collected using a gas syringe at
20.0 °C and 101 X 105.
Determine the limiting reactant and calculate the amount hydrogen gas produced in
(i) in mol and (ii) in cm3 under the stated conditions of temperature and pressure.
​

Answer 1

The magnesium is the limiting reagent and the moles of hydrogen are 3 × 10-3 and the volume is 72.356 × 10-2 cm³.

Given:

A student added 7.40 X 10-2g of magnesium ribbon to 15 cm3 of 2.00 mol dm-3

hydrochloric acid. The hydrogen gas produced was collected using a gas syringe at

20.0 °C and 101 X 105.

To Find:

The limiting reactant and calculate the amount of hydrogen gas produced in

(i) in mol and (ii) in cm3 under the stated conditions of temperature and pressure.

Solution:

To find the limiting reactant and calculate the amount of hydrogen gas produced in

(i) in mol and (ii) in cm3 under the stated conditions of temperature and pressure we will follow the following steps:

As we know,

Limiting reagents are calculated by taking the models of reactants and dividing them by their respective stoichiometry coefficients.

So,

$mg + 2hcl = mgcl2 + h2$

So,

Moles =

$\frac{given \: mass}{molar \: mass}$

The molar mass of magnesium = 24g

Given the mass of magnesium =

$7.40 \times {10}^{ - 2} g$

Stochiometric moles =

$\frac{moles}{stoichiometry} = \frac{moles}{1} = \frac{7.40 \times {10}^{ - 2} }{24} = 3 \times {10}^{ - 3}$

Now,

$moles \: of \: hydrogen \: = molarity \times volume \: in \: \: ml \: or\: \times {cm}^{3} \frac{1}{1000}$

$= 2 \times \frac{15}{1000} = 30 \times {10}^{ - 3}$

Stoichiometry of HCl = 2

So,

Stochiometric moles of HCl =

$\frac{30 \times {10}^{ - 3} }{2} = 15 \times {10}^{ - 3}$

So,

Magnesium stoichiometric moles are fewer so it is a limiting reagent.

Now,

Moles of H2 are determined by magnesium.

1 mole of magnesium gives 1 mole of hydrogen.

So,

$3 \times {10}^{ - 3}$

moles will give

$3 \times {10}^{ - 3}$

moles of hydrogen.

Now,

$p = 101 \times {10}^{5}$

temperature t = 20 + 273 = 293 K

R = 8.314

$n = 3 \times {10}^{ - 3}$

Now,

$v = \frac{nRt}{p} = \frac{3 \times {10}^{ - 3 } \times 8.314 \times 293}{101 \times {10}^{5} } {m}^{3} = 72.356 \times {10}^{ - 8} = 72.356 \times {10}^{ - 2} {cm}^{3}$

$(as \: 1 \: {m}^{3} = {10}^{6} {cm}^{3} )$

Henceforth, the magnesium is limiting reagent and the moles of hydrogen are 3 × 10-3 and the volume is 72.356 × 10-2 cm³.

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