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Paladinen [302]
3 years ago
11

What volume of nitrogen gas at STP would react with 37.2 g of magnesium to produce magnesium nitride

Chemistry
1 answer:
anastassius [24]3 years ago
6 0

Answer:

11.58 L of N₂

Explanation:

We'll begin by calculating the number of mole in 37.2 g of magnesium. This can be obtained as follow:

Mass of Mg = 37.2 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 37.2 / 24

Mole of Mg = 1.55 moles

Next, we shall write the balanced equation for the reaction. This is illustrated below:

3Mg + N₂ —> Mg₃N₂

From the balanced equation above,

3 moles of Mg reacted with 1 mole of N₂.

Therefore, 1.55 moles of Mg will react with = (1.55 × 1)/3 = 0.517 mole of N₂

Thus, 0.517 mole of N₂ is need for the reaction.

Finally, we shall determine the volume of N₂ needed for the reaction as follow:

Recall:

1 mole of a gas occupies 22.4 L at STP.

1 mole of N₂ occupied 22.4 L at STP.

Therefore, 0.517 mole of N₂ will occupy = 0.517 × 22.4 = 11.58 L at STP

Thus, 11.58 L of N₂ is needed for the reaction.

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1. What do gram-formula mass and atomic mass have in common? How are they different?
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The model of a “sea of electrons” helps explain ____.
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The correct answer is C.) metallic bonds

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A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

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