Trisilane (Si3H8) is a liquid with a density of 0.739 g cm-3. It reacts with oxygen to give silicon dioxide (SiO2) and water. Calculate the mass of water that would form if 26.9 cm3 of trisilane reacted completely with excess oxygen. _________g water
The Correct Answer and Explanation is:
To calculate the mass of water that would form when 26.9 cm³ of trisilane reacts completely with excess oxygen, we can follow these steps:
Step 1: Determine the number of moles of trisilane (Si₃H₈).
We are given the volume of trisilane (26.9 cm³) and its density (0.739 g/cm³). First, we calculate the mass of trisilane using the formula:Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}Mass=Density×VolumeMass of Si₃H₈=0.739 g/cm3×26.9 cm3=19.9 g\text{Mass of Si₃H₈} = 0.739 \, \text{g/cm}^3 \times 26.9 \, \text{cm}^3 = 19.9 \, \text{g}Mass of Si₃H₈=0.739g/cm3×26.9cm3=19.9g
Next, we calculate the number of moles of trisilane using its molar mass. The molar mass of Si₃H₈ is:Molar mass of Si₃H₈=(3×28.085 g/mol)+(8×1.008 g/mol)=88.258 g/mol\text{Molar mass of Si₃H₈} = (3 \times 28.085 \, \text{g/mol}) + (8 \times 1.008 \, \text{g/mol}) = 88.258 \, \text{g/mol}Molar mass of Si₃H₈=(3×28.085g/mol)+(8×1.008g/mol)=88.258g/mol
Now, calculate the moles of Si₃H₈:Moles of Si₃H₈=Mass of Si₃H₈Molar mass of Si₃H₈=19.9 g88.258 g/mol≈0.225 mol\text{Moles of Si₃H₈} = \frac{\text{Mass of Si₃H₈}}{\text{Molar mass of Si₃H₈}} = \frac{19.9 \, \text{g}}{88.258 \, \text{g/mol}} \approx 0.225 \, \text{mol}Moles of Si₃H₈=Molar mass of Si₃H₈Mass of Si₃H₈=88.258g/mol19.9g≈0.225mol
Step 2: Write the balanced chemical equation.
The balanced reaction between trisilane and oxygen to form silicon dioxide and water is:Si₃H₈+4 O₂→3 SiO₂+4 H₂O\text{Si₃H₈} + 4 \, \text{O₂} \rightarrow 3 \, \text{SiO₂} + 4 \, \text{H₂O}Si₃H₈+4O₂→3SiO₂+4H₂O
From this equation, we see that 1 mole of Si₃H₈ produces 4 moles of H₂O.
Step 3: Calculate the moles of water produced.
Since 1 mole of Si₃H₈ produces 4 moles of H₂O, the moles of water produced from 0.225 moles of Si₃H₈ is:Moles of H₂O=0.225 mol×4=0.900 mol\text{Moles of H₂O} = 0.225 \, \text{mol} \times 4 = 0.900 \, \text{mol}Moles of H₂O=0.225mol×4=0.900mol
Step 4: Calculate the mass of water produced.
The molar mass of water (H₂O) is:Molar mass of H₂O=(2×1.008 g/mol)+(1×15.999 g/mol)=18.016 g/mol\text{Molar mass of H₂O} = (2 \times 1.008 \, \text{g/mol}) + (1 \times 15.999 \, \text{g/mol}) = 18.016 \, \text{g/mol}Molar mass of H₂O=(2×1.008g/mol)+(1×15.999g/mol)=18.016g/mol
Now, calculate the mass of water:Mass of H₂O=Moles of H₂O×Molar mass of H₂O=0.900 mol×18.016 g/mol≈16.2 g\text{Mass of H₂O} = \text{Moles of H₂O} \times \text{Molar mass of H₂O} = 0.900 \, \text{mol} \times 18.016 \, \text{g/mol} \approx 16.2 \, \text{g}Mass of H₂O=Moles of H₂O×Molar mass of H₂O=0.900mol×18.016g/mol≈16.2g
Final Answer:
The mass of water formed is approximately 16.2 grams.
