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DIA [1.3K]
3 years ago
9

Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 mg + n2→ mg3n2 in a particular experiment

, a 10.1-g sample of n2 reacts completely. the mass of mg consumed is ________ g.
Chemistry
1 answer:
jekas [21]3 years ago
6 0
From the periodic table:
mass of Mg = 24 grams
mass of N = 14 grams

From the balanced equation we have:
3Mg + N2 ............> Mg3N2
we can see that:
3 moles of magnesium react with 2 moles of nitrogen.
This means that:
72 grams of magnesium are required to react with 28 grams of nitrogen.
Therefore, to calculate the amount of Mg needed to react with 10 grams of N2, all you have to do is cross multiplication as follows:
mass of Mg = (10*72) / 28 = 25.714 grams
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4 years ago
The average lung capacity of a human is 6.0L.
Darya [45]

Answer:

(a) 0.25 mol

(b) 0.11 mol

(c) 8.77 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol

(b)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol

(c)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol

8 0
4 years ago
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