Question

How many grams of water can be prepared from $ 10.10 $ grams of hydrogen gas and excess oxygen gas at standard conditions?

Answer 1

Hint :A molar mass ratio is used to transform a mass calculation to a substance amount. This quantity is expressed in particles such as atoms, molecules, or ions. It's the proportion of something's mass to the amount of particles that make it up.

Complete Step By Step Answer:
We have to begin by writing the equation of the reaction which is,
 $ {H_2} + {O_2} \to {H_2}O $
Now, we have to rewrite the balanced chemical equation for the reaction, so we get,
 $ 2{H_2} + {O_2} \to 2{H_2}O $
From the question, we know that oxygen is in excess, this means that hydrogen is the limiting reagent in the above reaction. So, hydrogen will decide the amount of product that is formed according to the mole ratio of the balanced equation.
 $ {n_{{H_2}}} = \dfrac{m}{{{M_r}}} $
where $ {n_{{H_2}}} $ is the amount of hydrogen in moles, $ m $ is the mass of hydrogen in grams and $ {M_r} $ is the molar mass of hydrogen (the mass of one mole of hydrogen) in $ g.mo{l^{ - 1}} $ .
We know that $ m $ which is the mass of hydrogen $ = $ $ 10.10g $ and
  $ {M_r} $ which is the molar mass of hydrogen $ = $ $ 2 \times 1 = 2g.mo{l^{ - 1}} $ [we got $ 2 $ since $ {H_2} $ has $ 2 $ molecules of $ H $ in it and molar mass of $ H $ is $ 1 $ ]
Now, on substituting the values, we get,
 $ {n_{{H_2}}} = \dfrac{m}{{{M_r}}} = \dfrac{{10.10g}}{{2g.mo{l^{ - 1}}}} = 5.05mol $
We have to find the mass of water required, so we use this formula,
 $
  {n_{{H_2}O}} = \dfrac{{{m_{{H_2}O}}}}{{{M_R}}} \\
   \Rightarrow {m_{{H_2}O}} = {n_{{H_2}O}} \times {M_R} \\
  $
Now, on substituting the values , we get,
 $ {M_R} $ which is the molar mass of water $ = $ $ (2 \times 1) + 16 = 18g.mo{l^{ - 1}} $ [we got $ 2 $ since $ {H_2} $ has $ 2 $ molecules of $ H $ in it and molar mass of $ H $ is $ 1 $ and the molar mass of $ O $ is $ 16 $ ]
 $ {m_{{H_2}O}} = 5.05 \times 18 = 90.9g $
Therefore, $ 90.9g $ grams of water can be prepared from $ 10.10 $ grams of hydrogen gas and excess oxygen gas at standard conditions.

Note :
A mole is represented as $ 6.0221 \times 1023 $ atoms, molecules, ions, or other chemical units. Because of the large number of atoms, molecules, and other particles in any substance, the mole is a simple unit to use.