70cm\(^3\) of hydrogen are sparked with 25cm\(^3\) of oxygen at S.T.P. The total volume of...

Explanation

The reaction between hydrogen and oxygen can be described by the following balanced chemical equation:

\(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\)

In this reaction, 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. This can also be interpreted in terms of volume, as 2 volumes of hydrogen gas react with 1 volume of oxygen gas to produce 2 volumes of water vapor at constant temperature and pressure (based on Avogadro's Law).

Given that 70 cm\(^3\) of hydrogen is sparked with 25 cm\(^3\) of oxygen at S.T.P. (Standard Temperature and Pressure), we can determine the limiting reactant and the volume of the residual gas.

To do this, we first need to compare the volume ratio of hydrogen to oxygen with the stoichiometric ratio from the balanced chemical equation. The stoichiometric ratio is 2:1 (hydrogen:oxygen), while the volume ratio is 70:25 or 2.8:1.

Since the volume ratio of hydrogen to oxygen is greater than the stoichiometric ratio, there is an excess of hydrogen, and oxygen is the limiting reactant.

From the stoichiometric ratio, 1 volume of oxygen reacts with 2 volumes of hydrogen. Therefore, the 25 cm\(^3\) of oxygen reacts with 50 cm\(^3\) of hydrogen. This leaves 20 cm\(^3\) of hydrogen unreacted (70 cm\(^3\) - 50 cm\(^3\)).

As a result, the total volume of the residual gas is 20 cm\(^3\) (unreacted hydrogen), which corresponds toOption A: 20 cm\(^3\).