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?
- it's given 45.0 mL of H2SO4 with the density of 1.84 g/mL, so the mass of H2SO4 is 45*1.84 = 82.8 (g) so converts to 0.845 moles of H2SO4
- the ratio between P4O10 and H2SO4 is 1 : 6, so the moles of P4O10 reacting is 0.141
so the mass of P4O10 reacts is 0.141*284 = 40.044 (g)
What volume of SO3(g) at STP is produced?
- the ratio between H2SO4 and SO3 is 1 : 1, so the moles of SO3 produced is 0.845
so the volume of P4O10 produced is 0.845*22.4 = 18.928 (L) in STP
A 200. mL sample of H3PO4 is titrated with 56.2 mL of 0.700M NaOH.
2 NaOH(s) + H3PO4(l) --> Na2HPO4(s) + 2H2O(l)
a) What is the molar concentration of H3PO4?
it's given 56.2 mL of 0.700M NaOH, so converts to 56.2/1000*0.7 = 0.04 moles of NaOH
- the ratio between NaOH and H3PO4 is 2 : 1, so the moles of H3PO4 needed is 0.02
- 0.02 moles H3PO4 in a 200 mL sample of H3PO4, that means the molar concentration of H3PO4 is 0.02/(200/1000) = 0.1(moles/L)
b) How many moles of Na2HPO4 are produced? (2 marks)
- the ratio between NaOH and Na2HPO4 is 2 : 1, so the moles of Na2HPO4 produced is 0.02
Looks good to me, except for 2B. Double check that one. The moles of H3PO4 are proportional to its conjugate base.
Nevermind. That one is correct as well.
Thank you !
Let's break down the information step by step:
a) To find the molar concentration of H3PO4, we need to use the given volume of NaOH solution and its concentration. The balanced equation tells us that the ratio between NaOH and H3PO4 is 2:1.
1. First, convert the given volume of NaOH solution from mL to L: 56.2 mL = 56.2/1000 L = 0.0562 L.
2. Next, calculate the number of moles of NaOH by multiplying its concentration by the volume: 0.7 M * 0.0562 L = 0.03934 moles of NaOH.
3. Since the ratio between NaOH and H3PO4 is 2:1, we know that 0.03934 moles of NaOH reacts with 0.03934/2 = 0.01967 moles of H3PO4.
4. Now, we need to calculate the molar concentration of H3PO4 by dividing the number of moles by the volume in liters: 0.01967 moles / (200 mL / 1000) L = 0.09835 moles/L.
Therefore, the molar concentration of H3PO4 is approximately 0.0984 moles/L.
b) To calculate the number of moles of Na2HPO4 produced, we refer to the balanced equation, which tells us that the ratio between NaOH and Na2HPO4 is 2:1. Since we have 0.01967 moles of NaOH, we can conclude that the same amount (0.01967 moles) of Na2HPO4 is produced.
Therefore, the number of moles of Na2HPO4 produced is 0.01967 moles.
H3PO4/H2SO4 problem is OK>
For M H3PO4 I get 0.098M
For mols Na2HPO4 I get 0.0196