When something is used only once and then thrown away it is said to be

7.32 moles of hydrogen reacts with 48.97 grams of nitrogen, how many moles of ammonia is produced?

faust18 [17]

3.5 moles of ammonia (NH₃) are produced

Explanation:

We have the following chemical reaction where hydrogen (H₂) reacts with nitrogen (N₂) to produce ammonia (NH₃):

3 H₂ + N₂ → 2 NH₃

number of moles = mass / molecular weight

number of moles of N₂ = 48.97 / 28 = 1.75 moles

We see from the chemical reaction that 1 mole of N₂ will react with 3 moles of H₂, so 1.75 moles of nitrogen will react with 3 × 1.75 = 5.25 moles of H₂. We have 7.32 moles of H₂, a quantity more of what is needed, so the limiting reactant is N₂.

Knowing this we devise the following reasoning:

if 1 mole of N₂ produces 2 moles of NH₃

then 1.75 moles of N₂ produces X moles of NH₃

X = (1.75 × 2) / 1 = 3.5 moles of NH₃

Learn more about:

limiting reactant

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7 0

3 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde

Eva8 [605]

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate

Thus, Ca2+ is present.

When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

8 0

3 years ago

An atom has eight protons and nine neutrons. draw the atomic structure of the atom

Cloud [144]

Draw eight circles and nine more in a big circle = this represents your nucleus
draw eight electrons (since electrons = protons in neutral elements) outside of it (2 in one ring, then 6 in a second ring)

7 0

2 years ago

Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.

kakasveta [241]

Answer:

The volume will be 89.6875 ml

Explanation:

So to count this we will use a single proportion.

0.0640 mol - 1000 ml

5.74×10−3 mol - x ml

x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml

6 0

3 years ago