Chemistry, 02.07.2019 09:30 hehefjf8610

What mass of nh3 in grams must be used to produce 1.81 tons of hno3 by the ostwald process, assuming an 80.0 percent yield in each step (1 ton=2000 lb; 1 lb= 453.6 g)?

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Answer from: aaleeyahprice

Mass of NH_3 used is 1730812.5 grams.

Explanation: Ostwald's process is the process of the production of nitric acid from ammonia. Steps for the process are:

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

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

$2NO_2(g)+H_2O(l)\rightarrow HNO_3(aq.)+HNO_2(aq.)$

Total mass of HNO_3 produced = 1.81 tons

Using conversion factor, we get

Total mass of HNO_3 produced = (1.81 × 2000 × 453.6)g = 1642032 g

Moles of HNO_3 is calculated by using the formula:

$\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Molar mass of nitric acid = 63 g/mol

$\text{Number of Moles}=\frac{1642032g}{63g/mol}=26064moles$

We are given that the percentage yield is 80 % in every step, then

Moles of NO_2 , theorectically (From Step-III) = 2 × 26064 = 52128 moles

Actual moles of $NO_2=52128\times \frac{100}{80}=65160moles$

Now, theoretical moles of NO (in Step-II) = Actual moles of NO_2

Theoretical moles of NO = 65160 moles

Actual moles of NO = $65160\times \frac{100}{80}=81450moles$

Now, theoretical moles of NH_3 (in Step-I) = Actual moles of NO

Theoretical moles of NH_3 = 81450 moles

Actual moles of NH_3 = $81450\times \frac{100}{80}=101812.5moles$

To calculate the mass of NH_3 , we use equation 1, we get

Molar mass of NH_3 = 17 g/mol

$\text{Mass of }NH_3\text{ used}=101812.5mol\times 17g/mol=1730812.5g$

Answer from: Quest

nanotechnology is a powerful tool in biomaterials research as it creates numerous possibilities for the engineering of functional molecules. nano-bio refers to projected ability to incorporate biomimicry into various materials at nanoscale.

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.