P4O10(s) + 6 H2SO4(l) 4 H3PO4(aq) + 6 SO3(g)
Pure H2SO4(l) has a density of 1.84 g/mL. If 45.0 mL of H2SO4(l) reacts:
What mass of P4O10 reacts?
45.0 mL X 1.84 g/mL X (1 mol/98.1 g) = 0.844 mol H2SO4
Moles P4O10 = 0.844 mol H2SO4 X (1 mol P4O10 / 6 mol H2SO4) = 0.141 mol P4O10
Mass P4O10 = 0.141 mol P4O10 X 283.9 g/mol = 39.9 g P4O10
What volume of SO3(g) at STP is produced? (2 marks)
0.844 mol H2SO4 X (6 mol SO3 / 6 mol H2SO4) = 0.844 mol SO3
At STP, 1 mol of any gas has a volume of 22.4 L. So,
Volume SO3 = 0.844 mol X (22.4 L/mol) = 18.9 L SO3
To determine the mass of P4O10 that reacts, we need to first calculate the number of moles of H2SO4 using its density and volume provided.
First, we calculate the number of grams of H2SO4:
Mass of H2SO4 = 45.0 mL * 1.84 g/mL = 82.8 g H2SO4
Next, we convert the grams of H2SO4 to moles using its molar mass:
Molar mass of H2SO4 = 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.1 g/mol
Moles of H2SO4 = 82.8 g / 98.1 g/mol = 0.844 mol
Now, we can use the balanced equation to determine the moles of P4O10 that react with the moles of H2SO4:
0.844 mol H2SO4 * (1 mol P4O10 / 6 mol H2SO4) = 0.141 mol P4O10
Finally, we can convert the moles of P4O10 to grams using its molar mass:
Molar mass of P4O10 = 4(31.00 g/mol) + 10(16.00 g/mol) = 283.9 g/mol
Mass of P4O10 = 0.141 mol * 283.9 g/mol = 39.9 g P4O10
Therefore, the mass of P4O10 that reacts is 39.9 g.
To determine the volume of SO3 gas produced at STP, we use the stoichiometry of the balanced equation:
0.844 mol H2SO4 * (6 mol SO3 / 6 mol H2SO4) = 0.844mol SO3
At STP, 1 mole of any gas occupies a volume of 22.4 L.
Therefore, the volume of SO3 gas produced is:
Volume SO3 = 0.844 mol * 22.4 L/mol = 18.9 L SO3
Therefore, the volume of SO3 gas produced at STP is 18.9 L.