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Vaselesa [24]
3 years ago
11

A balanced equation tells you the mole ratio of all reactants and products. Using the chemical reaction below how many moles of

NH3 are made when 3 moles of hydrogen gas are used?
2N2 + 3H2 ---> 2NH3
Chemistry
2 answers:
Hunter-Best [27]3 years ago
6 0
From  the  reaction  above    2  moles  of  N2  reacted  with  3  moles  of  H2  to  give  2  moles of   ammonia
THese  implies  that   the   mole  ratio  of  hydrogen  to  ammonia   is   3:2
Therefore  if   3  moles  of  hydrogen   were  used  the  moles  of  ammonia  is  therefore
(3 x2)/3=  2moles
vichka [17]3 years ago
5 0
<span>By using the mole ratio, we can determine that 2 moles of NH3 are made when 3 moles of hydrogen gas are present. The numbers in front of the chemicals tell us the relative amounts consumed and produced. Since there is a 3 in front of H2 and a 2 in front of NH3, this tells us that for every 3 moles of H2 gas used, 2 moles of NH3 are made.</span>
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What are two methods of writing a balanced formula?
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8 0
3 years ago
What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe with 2.1 g of O2?
Verizon [17]

Answer:

Fe is limiting, and it will produce .0188 mols of Fe2O3

Explanation:

after you convert both Fe and O2 to mols by using their molar mass, you see there is less Fe than O2 so that is your limiting reactant. To find the amount of Fe2O3 you devide the limiting reactant by it's coefeciant (4) then multiply it by the products coefficant (2). Let me know if you have any questions

8 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
2 years ago
QUESTIONS
vagabundo [1.1K]

Answer:

B.

Explanation:

3 0
3 years ago
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