Question

11:34 PM Tue May 26 425 < Question 17 of 48 Submit According to the balanced reaction below, calculate the moles of NH, that form when 4.2 mol of N,He completely reacts 3 NzH.M → NH3(g) + N (9) х STARTINO AMOUNT ANSWER RESET ADD FACTOR 42 16.8 1704 5.6 3.2 28.02 3 32.06 6.022 x 1023 mol N GN. HE ON mol NH, NH mol NH4 Tap here or pull up for additional resources

Answer 1

Answer = 5.6 mol of Ammoia is produced.

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$3N_{2}H_{4}(l)\rightarrow N_{2}(g)+4NH_{3}(g)$

According to the balanced Equation ,

3molesN2H41) = 4molesN H39)

Given that ,

4.2 mol of  $N_{2}H_{4}$ completely reacts , How much Ammonia is produced ?

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3molesN2H41) = 4molesN H39)

The number of moles of Ammoia produced when 4.2 mol of Hydrazine reacts completely--------

#moles of Ammonia =

$=4.2 moles of N_{2}H_{4}* \frac{4 moles of Ammonia}{3 moles of hydrazine}$

$=4.2 moles * \frac{4 moles }{3 moles}$

= 5.6 mol of Ammoia is produced when 4.2 moles of hydrazine completely reacted.