Question

1.) Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas.
2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)

What mass of H2(g) is required from the reaction of 0.75 g of Al(s) with excess hydrochloric acid?


2.) The reaction of coal and water at a high temperature produces a mixture of hydrogen and carbon monoxide gases. This mixture is known as synthesis gas (or syngas). What mass of water is required for the formation of 175 grams of carbon monoxide?
C(s) + H20(g) -> H2(g) + CO(g)


3.) Nitroglycerine decomposes violently according to the chemical equation below. What mass of carbon dioxide gas is produced from the decomposition of 5.00 g C3H5(NO3)3?
__C3H5(NO3)3 (l) -> __CO2 (g) + __N2 (g) + __H2O (g) + __O2 (g)

Answer 1

Concepts and reason

The mass of products produced from the reactants depends on the amount of limiting reactant and the moles of reactants and products present in a balanced chemical equation. Limiting reactant is the substance which is present in lesser amount than required to react with the other reactant.

Fundamentals

The balanced chemical equation is the equation in which the numbers of atoms of different elements present on the left hand side of the equation is equal to that of in right hand side.

1)

The reaction is as follows:

${\rm{2Al}}\left( s \right){\rm{ + }}\,{\rm{6HCl}}\left( {aq} \right) \to {\rm{2AlC}}{{\rm{l}}_3}\left( {aq} \right){\rm{ + }}\,{\rm{3}}{{\rm{H}}_2}\left( g \right)$

Here, 2 moles of ${\rm{Al}}\left( s \right)$gives 3 moles of${{\rm{H}}_2}\left( g \right)$. There is excess of hydrochloric acid and$0.75\,{\rm{g}}\,\,{\rm{of}}\,\,{\rm{Al}}\left( s \right)$. Hence, the limiting reagent is${\rm{Al}}\left( s \right)$.

Calculate the mass of ${{\rm{H}}_2}\left( g \right)$formed as follows:

$\begin{array}{l}\\{\rm{Mass}}\,\,{\rm{of}}\,\,{{\rm{H}}_2}\, = \,\,0.75\,{\rm{g}}\,{\rm{of}}\,{\rm{Al}}\left( s \right) \times \frac{{1\,{\rm{mol}}\,{\rm{Al}}}}{{26.98\,{\rm{g}}\,{\rm{Al}}}} \times \frac{{3\,{\rm{mol}}\,{{\rm{H}}_2}}}{{2\,{\rm{mol}}\,{\rm{Al}}}} \times \frac{{2.00\,{\rm{g}}\,{{\rm{H}}_2}}}{{{\rm{1}}\,{\rm{mol}}\,{{\rm{H}}_2}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,0.083\,{\rm{g}}\\\end{array}$

2)

The reaction is as follows:

${\rm{C}}\left( s \right){\rm{ + }}\,{{\rm{H}}_2}{\rm{O}}\left( g \right) \to {{\rm{H}}_2}\left( g \right){\rm{ + }}\,{\rm{CO}}\left( g \right)$

Here, 1 mole of ${{\rm{H}}_2}{\rm{O}}$gives 1 mole of${\rm{CO}}\left( g \right)$.

Calculate the mass of ${{\rm{H}}_2}{\rm{O}}\left( g \right)$formed as follows:

$\begin{array}{l}\\{\rm{Mass}}\,\,{\rm{of}}\,\,{{\rm{H}}_2}{\rm{O}}\, = \,\,175\,{\rm{g}}\,{\rm{of}}\,{\rm{CO}} \times \frac{{1\,{\rm{mol}}\,{\rm{CO}}}}{{28.01\,{\rm{g}}\,{\rm{CO}}}} \times \frac{{1\,{\rm{mol}}\,{{\rm{H}}_2}{\rm{O}}}}{{1\,{\rm{mol}}\,{\rm{CO}}}} \times \frac{{18.00\,{\rm{g}}\,{{\rm{H}}_2}{\rm{O}}}}{{{\rm{1}}\,{\rm{mol}}\,{{\rm{H}}_2}{\rm{O}}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,112.5\,{\rm{g}}\\\end{array}$

3)

The reaction is as follows:

${{\rm{C}}_3}{{\rm{H}}_5}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\left( l \right) \to {\rm{C}}{{\rm{O}}_2}\left( g \right){\rm{ + }}{{\rm{N}}_2}\left( g \right){\rm{ + }}\,\,{{\rm{H}}_2}{\rm{O}}\left( g \right) + {{\rm{O}}_2}\left( g \right)$

The balanced reaction is as follows:

${\rm{4}}{{\rm{C}}_3}{{\rm{H}}_5}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\left( l \right) \to 12{\rm{C}}{{\rm{O}}_2}\left( g \right){\rm{ + }}\,{\rm{6}}{{\rm{N}}_2}\left( g \right){\rm{ + }}\,10\,{{\rm{H}}_2}{\rm{O}}\left( g \right) + {{\rm{O}}_2}\left( g \right)$

Calculate the mass of ${\rm{C}}{{\rm{O}}_2}\left( g \right)$formed as follows:

$\begin{array}{l}\\{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{C}}{{\rm{O}}_2}\, = \,\,5.00\,{\rm{g}}\,{\rm{of}}\,{{\rm{C}}_3}{{\rm{H}}_5}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2} \times \frac{{1\,{\rm{mol}}\,{{\rm{C}}_3}{{\rm{H}}_5}{{\left( {{\rm{N}}{{\rm{O}}_3}} \right)}_2}}}{{227.06\,{\rm{g}}\,{{\rm{C}}_3}{{\rm{H}}_5}{{\left( {{\rm{N}}{{\rm{O}}_3}} \right)}_2}}}\,\,\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \frac{{12\,{\rm{mol}}\,{\rm{C}}{{\rm{O}}_2}}}{{4\,{\rm{mol}}\,{{\rm{C}}_3}{{\rm{H}}_5}{{\left( {{\rm{N}}{{\rm{O}}_3}} \right)}_2}}} \times \frac{{44.02\,{\rm{g}}\,{\rm{C}}{{\rm{O}}_2}}}{{{\rm{1}}\,{\rm{mol}}\,{\rm{C}}{{\rm{O}}_2}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,2.91\,{\rm{g}}\\\end{array}$

Ans:

Therefore, the mass of ${{\rm{H}}_2}\left( g \right)$ is$0.083\,{\rm{g}}$.

Therefore, the mass of ${{\rm{H}}_2}{\rm{O}}\left( g \right)$required is$112.5\,{\rm{g}}$.

Therefore, the mass of ${\rm{C}}{{\rm{O}}_2}\left( g \right)$produced is$2.91\,{\rm{g}}$.