Chemistry, 07.04.2020 01:35 harodkdc7910

The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2 in the process. The P4O6 can be treated with water to form H3PO3, which forms PH3 and H3PO4 when heated. Find the mass of PH3 that forms from the reaction of 1.95 g NH3.

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Answer from: heyitseddie06

Answer : The mass of PH_3 produced from the reaction is, 0.651 grams.

Explanation :

The chemical reactions used are:

(1) $4NH_3+5O_2\rightarrow 4NO+6H_2O$

(2) $6NO+P_4\rightarrow P_4O_6+3N_2$

(3) $P_4O_6+6H_2O\rightarrow 4H_3PO_4$

(4) $4H_3PO_4\rightarrow PH_3+3H_3PO_4$

First we have to calculate the moles of NH_3

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mol

$\text{Moles of }NH_3=\frac{1.95g}{17g/mol}=0.115mol$

Now we have to calculate the moles of

From the balanced chemical reaction 1, we conclude that:

As, 4 moles of NH_3 react to give 4 moles of

So, 0.115 moles of NH_3 react to give 0.115 moles of

Now we have to calculate the moles of P_4O_6

From the balanced chemical reaction 2, we conclude that:

As, 6 moles of react to give 1 moles of P_4O_6

So, 0.115 moles of react to give $\frac{0.115}{6}=0.0192$ moles of P_4O_6

Now we have to calculate the moles of H_3PO_4

From the balanced chemical reaction 3, we conclude that:

As, 1 moles of P_4O_6 react to give 4 moles of H_3PO_4

So, 0.0192 moles of P_4O_6 react to give $0.0192\times 4=0.0768$ moles of H_3PO_4

Now we have to calculate the moles of PH_3

From the balanced chemical reaction 4, we conclude that:

As, 4 moles of H_3PO_4 react to give 1 moles of PH_3

So, 0.0768 moles of H_3PO_4 react to give $\frac{0.0768}{4}=0.0192$ moles of PH_3

Now we have to calculate the mass of PH_3

$\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3$

Molar mass of PH_3 = 33.9 g/mole

$\text{ Mass of }PH_3=(0.0192moles)\times (33.9g/mole)=0.651g$

Therefore, the mass of PH_3 produced from the reaction is, 0.651 grams.

Answer from: Quest

They are all solids.

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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The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2...