Question

when the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor.

part d

how many grams of oxygen are required to produce 48.2 g of water vapor?
express your answer with the appropriate units.

incorrect; try again; 4 attempts remaining
to determine the mass of oxygen required to for the given quantity of water vapor, start by converting the moles of product, you can use the coefficients in the balanced chemical equation to develop a mole-mole convert the number of moles of oxygen that react to grams using its molar mass.

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(2H_2S + 3O_2
ightarrow 2SO_2+2H_2O\)

Step2: Convert mass of \(H_2O\) to moles

The molar mass of \(H_2O\) is \(M = 18.02\space g/mol\). Using the formula \(n=\frac{m}{M}\), for \(m = 48.2\space g\) of \(H_2O\), \(n_{H_2O}=\frac{48.2\space g}{18.02\space g/mol}\approx 2.675\space mol\)

Step3: Use mole - mole ratio

From the balanced equation, the mole - mole ratio of \(O_2\) to \(H_2O\) is \(\frac{3\space mol\space O_2}{2\space mol\space H_2O}\). So \(n_{O_2}=n_{H_2O}\times\frac{3}{2}\). Substituting \(n_{H_2O} = 2.675\space mol\), \(n_{O_2}=2.675\space mol\times\frac{3}{2}=4.0125\space mol\)

Step4: Convert moles of \(O_2\) to mass

The molar mass of \(O_2\) is \(M = 32.00\space g/mol\). Using \(m = n\times M\), \(m_{O_2}=4.0125\space mol\times32.00\space g/mol = 128.4\space g\)

Answer:

\(128\space g\)