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Monica [59]
3 years ago
12

Phosphorus pentachloride reacts with water to form hydrochloric acid and phosphoric acid. How many total moles of acid are forme

d when starting with 4.5 g of PCl5 and excess H2O?
Chemistry
1 answer:
klio [65]3 years ago
3 0

Answer:

0,13 moles of acid are produced

Explanation:

The reaction of the problem is:

PCl₅ + 4H₂O → 5HCl + H₃PO₄

Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).

The molecular mass of PCl₅ is:

1P = 30,97g/mol + 5Cl = 5×35,45g/mol = <em>208,24 g/mol</em>

That means 4,5g of PCl₅ are:

4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:

0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =

<em>0,13 moles of acid are produced</em>

I hope it helps!

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Answer:

B??? sorry if im wrong

Explanation:

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3 years ago
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Carbon disulfide is formed by the reaction of coke (carbon) with sulfur dioxide. how many moles of cs2 will be generated if 8.0
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3 years ago
A cube has a height of 8 cm and a mass of 457 g. What is its density?
ZanzabumX [31]

Answer:

the answer is b,  0.89

Explanation:

A cube has a height of 8 cm and a mass of 457 g. What is its density?

a. 233,984 g/cm

b.  0.89g/cm3

c. 1.12 g/cm3

Density = mass/volume

the volume of the cube is 8X8X8=512cm3

the mass is 457 gm

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3 0
2 years ago
A chemist requires 0.802 mol Na2CO3 for a reaction. How many grams does this correspond to?
Komok [63]

Answer:

Ok:

Explanation:

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7 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
3 years ago
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