If my boyfriend asked me out on October 19 when will my one month be ?

Limiting reactants would appreciate the help

Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.

Your Problem:

Given: 3Ba + N₂ => Ba₃N₂

22.6g 4.2g ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ => 4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

Divide each mole value by respective coefficient ... smallest value is Limiting Reactant.

Barium => 0.165/3 = 0.055 <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

<em>Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant. </em>

3Ba + N₂ => Ba₃N₂ (3:1 rxn ratio for Ba:Ba₃N₂)

moles 0.165mole 0.150mole 1/3(0.165)mole = 0.055mole Ba₃N₂

= 0.055mol(440g/mol) Ba₃N₂

= 24.2 grams Ba₃N₂ (as based

upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) = moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

5 0

3 years ago

How many grams of NaF form when .5 mol of HF reacts with excess Na2SiO3?

IgorLugansk [536]

Answer:

5.25g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O

From the balanced equation above,

8 moles of HF reacted to produce 2 moles of NaF.

Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.

Next, we shall convert 0.125 mole of NaF to grams.

This is illustrated below:

Mole of NaF = 0.125 mole

Molar mass of NaF = 23 + 19 = 42g/mol

Mass of NaF =..?

Mass = mole x molar mass

Mass of NaF = 0.125 x 42

Mass of NaF = 5.25g

Therefore, 5.25g of NaF is produced from the reaction.

7 0

2 years ago

N2+3H2 → 2NH3

s2008m [1.1K]

Explanation:

N2 (g) + H2 (g) gives out NH3 (g)

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give NH3 a new stoichiometric coefficient.)

N2 (g) + 3H2 (g) gives out 2NH3 (g)

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one mol of anything is equal in quantity to one mol of anything else because the comparison is made in the units of mols.

So what do we do? Convert to

mols (remember the hint?).

28g N2 × 1 mol N2/ 2 × 14.007) g N2

= 0.9995 mol N2

At this point you don't even need to calculate the number of mols of H2 . Why? Because H2 is about 2 g/mol, which means we have over 10 mols of H2. We have 1 mol N2, and we need three times as many mols of H2 as we have

N2.

After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.

Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:

1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3

= 34.0444 g NH3

Hope it helpz~

4 0

2 years ago

what is the percent yield of NaCl if 31.0 g of CuCl reacts with excess NaNo3 to produce 21.2 g of NaCl

olchik [2.2K]

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

Balanced Equation: CuCl + NaNO₃   → NaCl + CuNO₃

Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol

the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,

then moles of NaCl = 0.31 mol

Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g

⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.

Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %

NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.

4 0

3 years ago

What is the overall energy transformation in a coal-fired power plant?.

polet [3.4K]

Answer: When coal is burned, chemical potential energy is transformed into thermal energy, light energy, and sound energy. Only the thermal energy is used for electricity production. Light and sound energy dissipate into the environment, immediately reducing efficiency.

Hope this helps.

7 0

2 years ago