Question

Calculate how many moles of sulphur dioxide will be evolved when 5moles of sulphur is reacted with oxygen

Answer 1

Answer:

• 5 Moles of Sulphur Dioxide will be released.

Given:

1. 5 moles of Sulphur reacts with Oxygen.

Explanation:

$\rule{300}{1.5}$

From the Reaction we Have.

$\large\bigstar \: {\boxed{\tt S_{(s)} + O_{2_{(g)}} \xrightarrow{} SO_{2_{(g)}}}}$

Here we Can see,

$\large{\tt \text{1 mole of Sulphur} \longrightarrow \text{1 mole of} \: SO_2}$

Now,

$\large{\tt \text{5 mole of Sulphur} \longrightarrow 5 \times \text{1mole of} \: SO_2}$

$\large{\underline{\boxed{\tt \text{5 mole of Sulphur} \longrightarrow \text{5 mole of} \: SO_2}}}$

∴ 5 Moles of Sulphur Dioxide will be released.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Additional information:-

$\large\boxed{\begin{minipage}{7 cm} \underline{\tt Important \: Formulas:-} \\ \\ \tt Moles (n) = \dfrac{Weight}{Molar \: Mass} \\ \\ \tt Moles (n) = \dfrac{No. \: of \: Particles}{Avogadro's \: Number} \\ \\ \tt Moles (n) = \dfrac{Volume}{22.4\: L} \\ \\ \tt Molarity = \dfrac{Moles \: of \: Solute}{Volume \: of \: Solution(L)} \\ \\ \tt Molality = \dfrac{Moles \: of \: Solute}{Mass \; of \: Solvent} \\ \\ \tt Normality = \dfrac{Gram \: Equivalents}{Volume \: of \: Solution (L)} \end{minipage}}$

$\rule{300}{1.5}$