Ask question

Ask question

All categories

ICE Princess25 [194]

2 years ago

9

Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg N2 → Mg3N2 In a particular experiment,

a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.

1 answer:

vlada-n [284]2 years ago

5 0

Answer:

The mass of Mg consumed is 21.42g

Explanation:

The reaction is

$3Mg+N_{2}-->Mg_{3}N_{2}$

As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.

as given that mass of nitrogen reacted = 8.33g

So moles of nitrogen reacted = $\frac{mass}{molarmass}=\frac{8.33}{28}=0.2975mol$

moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol

Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g

You might be interested in

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

2 years ago

What volume, in ml, of a 0.2089 m ki solution contains enough ki to react exactly with the cu(no3)2 in 43.88 ml of a 0.3842 m so

Triss [41]

The reaction is given as: $2Cu(NO_{3})_{2}+4KI\rightarrow 2CuI+I_{2}+4KNO_{3}$ Here, two moles of copper nitrate reacts with four moles of potassium iodide to give two moles of copper iodide, one mole of iodine and four moles of potassium nitrate. First, calculate the number of moles of copper nitrate. Number of moles is equal to the product of molarity and volume of solution in litre. Number of moles = $0.3842 M\times 0.04388 L$ (1 L =1000 mL)

= $0.016858696 mole$

Copper nitrate requires = $0.016858696 \times \frac{4}{2}$ mole of potassium iodide

= $0.033717392 mole$ of potassium iodide

Volume of solution in litre = $\frac{number of moles}{Molarity}$

Thus, volume of potassium iodide is = $\frac{0.033717392}{0.2089}$

= $0.1614 L$

1 L =1000 mL

Volume of potassium iodide in mL = $161.4 mL$

Hence, $161.4 mL$ 0.2089 M potassium iodide consist of sufficient potassium iodide to react with copper nitrate in 3.88 mL of a 0.3842 M solution of copper nitrate .

7 0

2 years ago

Which of the following is a class 3 surgical candidate

nataly862011 [7]

Ikr behehbenekebe sgwhebejebeb

5 0

2 years ago

An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine

nekit [7.7K]

Answer:

The molecular formula of the compound is $C_{7}H_{6}O_{2}$ .

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

$m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{C} = 84.067\,g$

Hydrogen

$m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{H} = 6.057\,g$

Oxygen

$m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{O} = 31.995\,g$

Now, the number of moles ( $n$ ), measured in moles, of each element are calculated by the following expression:

$n = \frac{m}{M}$

Where:

$m$ - Mass of the element, measured in grams.

$M$ - Molar mass of the element, measured in grams per mol.

Carbon ( $m_{C} = 84.067\,g$ , $M_{C} = 12.011\,\frac{g}{mol}$ )

$n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }$

$n = 7$

Hydrogen ( $m_{H} = 6.057\,g$ , $M_{H} = 1.008\,\frac{g}{mol}$ )

$n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }$

$n = 6$

Oxygen ( $m_{O} = 31.995\,g$ , $M_{O} = 15.999\,\frac{g}{mol}$ )

$n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }$

$n = 2$

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

$C_{7}H_{6}O_{2}$

8 0

2 years ago

What is the total E associated with one mole of photons (a unit known as the Einstein) of 3.91x1019 Hz?

sashaice [31]

Answer:

$E=1.56\times 10^{10}J$

Explanation:

The<em> energy of a photon</em>, E, can be calculated with the Planck-Einstein equation:

$E=hf$

Where:

h is Planck's constant 6.626×10⁻³⁴ J.s, and

f is the frequency of the photon or electromagnetic radiation.

Substituting with your data:

$E=6.626\times 10^{-34}J.s\times 3.91\times 10^{19}s^{-1}=2.5908\times 10^{-14}J$

Now multiply by Avogadro's number to obtain the energy of one mole of photons:

$E=2.5908\times 10^{-14}J\times 6.022\times 10^{23}=1.56\times 10^{10}J$

8 0

2 years ago