Chemistry, 14.07.2019 04:00 rainingsouls

How many moles of ca3 (po4)2 can be formed when 1.9 moles of caco3 react with 2.7 moles of fepo4?

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Answer from: adenmoye0910

0.633 moles

Explanation:-

The first step to solving this problem is to write the balanced chemical equation.

The balanced chemical equation for this reaction is

3 CaCO3 + 2 FePO4 = Ca3(PO4)2 + Fe2(CO3)3

From the balanced chemical equation, we can see

3 moles of CaCO3 react with 2 moles of FePO4.

In the question we are told there are 1.9 moles of CaCO3.

So 1.9 moles of CaCO3 react with (2 moles x 1.9 moles )/ 3 moles= 1.267 moles of FePO4.

The question tells us there are 2.7 moles of FePO4.

So there is excess FePO4.

Thus our limiting reagent is CaCO3. It will determine how much Ca3(PO4)2 is formed.

From the balanced chemical equation, we can see

3 moles of CaCO3 gives 1 mole of Ca3(PO4)2

1.9 moles of CaCO3 gives (1 moles x 1.9 moles ) / 3 moles= 0.633 moles of Ca3(PO4)2.

0.633 moles of Ca3 (PO4)2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4

Answer from: Quest

Forgot the question hahaha

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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