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3 years ago

Determine the number of moles of 35.4 grams of Magnesium.

Chemistry

1 answer:

dezoksy [38]3 years ago

5 0

Answer:

1.46 moles

Explanation:

You need to multiply 35.4g by Magnesium's molar mass.

Molar mass =1 mole/24.305g

(35.4g)(1 mole Mg/24.305g)= 1.46 moles <- rounded to 3 significant figures

As you can see, grams cancel out and you're left with moles.

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In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li = 6.94 g/mol

Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g

5 0

3 years ago

Naturally occurring boron has an atomic weight of 10.811. Its principal isotopes are 10B and 11B. Part A What is the abundance (

Darya [45]

Answer:

10B has 18.9%

11B has 81.1%

Explanation:Please see attachment for explanation

4 0

4 years ago

Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all char

Sonbull [250]

Answer:

2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓

Ksp = [2s]² . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp

2 s s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]² . [s] → 4s³

3 0

3 years ago

Formulae for calculating number of moles

Tresset [83]

You would use this number, 6.02×1023 (Avogadro's number) to convert from particles, atoms, or molecules to moles. Whenever you go to the mole, divide by Avogadro's number. When you go to the unit from moles, multiply by Avogadro's number.

4 0

3 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?

BigorU [14]

Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.

6 0

4 years ago

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