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Murljashka [212]
3 years ago
11

6 Li + N2 --> 2 Li3N How much N2 is needed to make 1.34 moles of Li3N?

Chemistry
1 answer:
Alex777 [14]3 years ago
4 0

Answer: 0.67 moles

Explanation:

6 Li(s) + N2(g) ---> 2 Li3N(s)

The balanced reaction above shows that 6 moles of Lithium, (Li) added to 1 mole of Nitrogen gas, (N2) yields 2 moles of Lithium nitrite, (Li3N)

I.e 1 mole of N2 = 2 moles of Li3N

Since the number of N2 needed to make 1.34 moles of Li3N is unknown, it is represented by Y

So, 1 mole of N2 = 2 moles of Li3N

Y = 1.34 moles of Li3N

To get the value of Y, cross multiply

1.34 x 1 = Y x 2

1.34 = 2Y

To get the value of Y divide both sides by 2

1.34/2 = 2Y/2

0.67 = Y

Thus, 0.67 moles of N2 is needed to make 1.34 moles of Li3N.

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Neporo4naja [7]

Answer: There are 0.056 moles present in 3.45 g of Na_{2}O.

Explanation:

Given : Mass = 3.45 g

Moles is the mass of substance divided by its molar mass.

Hence, moles of Na_{2}O (molar mass = 61.98 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{3.45 g}{61.98 g/mol}\\= 0.056 mol

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What is the concentration in mol L-l of a 12% solution of tetrahydrate magnesium chloride (MgCl2)?
madreJ [45]

The concentration in mol L-l (M) = 0.717

<h3>Further explanation</h3>

Given

12% solution of tetrahydrate magnesium chloride (MgCl₂)

Required

The concentration

Solution

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MW = Ar Mg+2. Ar Cl+8. Ar H + 4. Ar O

MW=24.31 + 2 x 35.45 + 8 x 1.01 + 4 x 16

MW=24.31+70.9+8.08+64

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12% solution = 12 % m/v = 12 g in 100 ml solution

For 1 L solution :

\tt \dfrac{1}{0.1}\times 12~g=120~g

The concentration in g/L = 120 g/L

Convert grams to moles :

\tt mol=\dfrac{120}{167.29}=0.717

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