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djverab [1.8K]
2 years ago
13

5F248+ 2NH3(8) ► NF418 + 6HF

Chemistry
1 answer:
mylen [45]2 years ago
6 0

Answer:

14.2L at STP

Explanation:

Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:

<em>Moles NH3 -Molar mass: 17.031g/mol-</em>

3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3

<em>Moles HF:</em>

0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF

<em>Volume HF</em>

PV = nRT; V = nRT/P

<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>

Replacing:

V = 0.634moles*0.082atmL/molK*273.15K / 1atm

V = 14.2L at STP

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Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

7 0
3 years ago
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Answer:

the energy of the products is less than the energy of the reactants.

Explanation:

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Answer:

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Explanation:

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