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vampirchik [111]

4 years ago

5

If 12.85 g of chromium metal is reacted with 10.72 g of phosphoric acid, then what is the maximum mass in grams of chromium(III)

phosphate that can be produced? 2Cr(s) + 2H3PO4(aq) → 2CrPO4(s) + 3H2(g)

1 answer:

RUDIKE [14]4 years ago

7 0

<h3>Answer:</h3>

16.02 g

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of chromium as 12.85 g
Mass of phosphoric acid as 10.72 g

We are require to calculate the maximum mass of chromium (III) phosphate that can be produced.

The equation for the reaction is;

2Cr(s) + 2H₃PO₄(aq) → 2CrPO₄(s) + 3H₂(g)

<h3>Step 1: Determining the number of moles of Chromium and phosphoric acid</h3>

Moles = Mass ÷ Molar mass

Molar mass of chromium = 52.0 g/mol

Moles of Chromium = 12.85 g ÷ 52.0 g/mol

= 0.247 moles

Molar mass of phosphoric acid = 97.994 g/mol

Moles of phosphoric acid = 10.72 g ÷ 97.994 g/mol

= 0.109 moles

<h3>Step 2: Determine the rate limiting reactant </h3>

From the equation, 2 moles of chromium reacts with 2 moles of phosphoric acid.
Therefore, 0.247 moles of Chromium will require 0.247 moles of phosphoric acid, but we only have 0.109 moles.

This means chromium is in excess and phosphoric acid is the rate limiting reagent.

<h3>Step 3: Moles of Chromium(III) phosphate</h3>

From the equation, 2 moles of phosphoric acid reacts to yield 2 moles of chromium (III) phosphate.
Therefore, Moles of Chromium (III) phosphate = Moles of phosphoric acid

Hence; moles of CrPO₄ = 0.109 moles

<h3>Step 4: Maximum mass of CrPO₄ that can be produced</h3>

We know that, mass = Moles × Molar mass

Molar mass of CrPO₄ = 146.97 g/mol

Thus,

Mass of CrPO₄ = 0.109 moles × 146.97 g/mol

= 16.02 g

Thus, the maximum mass of CrPO₄ that can be produced is 16.02 g

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Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

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Explanation:

<u>Part I :</u>

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

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This means, 1 mole of gold(Au) contain = 196.96 grams

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<u>Part II :</u>

Density of Gold = $19.32 g/cm^{3}$

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Volume of the gold bar = $6.00\times 4.25\times 2.00$

Volume of the gold bar = 51 $cm^{3}$

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natali 33 [55]

Answer:

I would say is B but I do t really know

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1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

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Answer:

1) The new pressure of the gas is 500 kilopascals.

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Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

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$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

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The new pressure of the gas is 500 kilopascals.

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$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

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$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

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