Answer:
a) mass of H₂O produced = 4.0665 * 18 = 73.197 g
b)mass of H₂O produced = 0.214 * 18 = 3.852 g
c) Oxygen gas, O₂, is the limiting reactant, since the number of moles available for reaction is far smaller and the mass of water produced from it is smaller too.
<em>The reaction is the combustion of hydrogen gas to form water vapor. The question is not complete. A related question is given below:</em>
<em>Consider the following balanced equation: </em>
<em>2H2 + O2 --------> 2H2O </em>
<em>If you start with 8.133 g of H2 and 3.425 g of O2, find the following:
</em>
<em>a) With excess O2, what mass (grams) of H2O would be produced by the H2?
</em>
<em>b) With excess H2, what mass (grams) of H2O would be produced by the O2?
</em>
<em>c) What is the chemical formula for the limiting reactant?</em>
Explanation:
Equation of reaction: 2H₂ + O₂ ---> 2H₂O
From the equation of reaction, 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce two moles of water
Molar mass of hydrogen gas, H₂ = 2 g/mol; molar mass of oxygen gas, O₂ = 32 g/mol; molar mass of H₂O = 18 g/mol
a) number of moles of hydrogen in 8.133 g = 8.133 g/ 2g/mol = 4.0665 moles
mole ratio of H₂ to H₂O is 1:1, therefore, 4.0665 moles of H₂O will be produced
mass of H₂O produced = 4.0665 * 18 = 73.197 g
b) number of moles of oxygen in 3.425 g = 3.425 g /32 g/mol = 0.107 moles
mole ratio of O₂ to H₂O is 1:2, therefore, 0.107 * 2 moles of H₂O will be produced = 0.214 moles of H₂O
mass of H₂O produced = 0.214 * 18 = 3.852 g
c) number of moles of hydrogen in 8.133 g = 4.0665 moles
number of moles of oxygen in 3.425 g = 0.107 moles
mole ratio of H₂ to O₂ = 4.0665/0.107 = 38 : 1
Oxygen gas, O₂, is the limiting reactant, since the number of moles available for reaction is far smaller and the mass of water produced from it is smaller too.
