All categories

2 years ago

Consider a 175.7 g sample of the compound manganese(IV) perchlorate.

Chemistry

1 answer:

kramer2 years ago

6 0

Answer:

6.21 moles O

Explanation:

To find the moles of oxygen, you need to (1) convert grams Mn(ClO₄)₄ to moles Mn(ClO₄)₄ (via molar mass) and then (2) convert moles Mn(ClO₄)₄ to moles O (via mole-to-mole ratio from formula subscripts). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

Molar Mass (Mn(ClO₄)₄): 452.74 g/mol

1 Mn(ClO₄)₄ = 1 Mn and 4 Cl and 16 O

175.7 g Mn(ClO₄)₄ 1 mole 16 moles O
--------------------------- x ------------------- x --------------------------- = 6.21 moles O
452.74 g 1 mole Mn(ClO₄)₄

You might be interested in

The fluid in a car battery has a volume of 125 mL and a mass of 155 g what is the density

mihalych1998 [28]

Answer:

derhsaw ₹ lagega samjha kya bete

4 0

3 years ago

Read 2 more answers

Write the formula for the compound formed between aluminum and nitrogen

Yuki888 [10]

The aluminum formula for the compound formed between aluminum and nitrogen is AlN.

4 0

3 years ago

A piece of an unknown metal has a volume of 19.8 ml and a mass of 210.0 grams. The density of the metal is g/mL A piece of the s

leva [86]

Answer: The density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL

Explanation:

To calculate the density of unknown metal, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ ......(1)

Volume of unknown metal = 19.8 mL

Mass of unknown metal = 210.0 g

Putting values in equation 1, we get:

$\text{Density of unknown metal}=\frac{210.0g}{19.8mL}\\\\\text{Density of unknown metal}=10.60g/mL$

The density of the metal remains the same.

Now, calculating the volume of unknown metal, using equation 1, we get:

Density of unknown metal = 11.45 /mL

Mass of unknown metal = 86.0 g

Putting values in above equation, we get:

$10.60g/mL=\frac{86.0g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.11mL$

Hence, the density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL

5 0

3 years ago

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.4HCl(aq)+MnO2(s)⟶Mn

mario62 [17]

Answer:

HCl is limiting reactant

Theoretical yield: 23.8g Cl₂

Actual yield: 17.6g C₂

Explanation:

Based on the reaction:

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.

To find the limiting reactant you must know the moles of each reactant and knowing that 4 moles of HCl reacts per mole of MnO₂ you can sikve this problem, thus:

Moles HCl (Molar mass: 36.46g/mol): 48.9g ₓ (1mol / 36.46g/mol) =

1.341 moles HCl

Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =

0.424 moles MnO₂

For a complete reaction of 0.424 moles of MnO₂ you require:

0.424moles MnO₂ ₓ (4 moles HCl / 1 mole MnO₂) = 1.696 moles of HCl.

As you have just 1.341 moles of HCl. HCl is limiting reactant.

Theoretical yield means, in the reaction, that 4 moles of HCl will produce 1 mole of Cl₂. As moles of HCl are 1.341:

1.341 moles HCl ₓ (1 mole Cl₂ / 4 moles HCl) = 0.33525 moles Cl₂

In grams (Molar mass Cl₂: 70.9g/mol):

Theoretical yield: 0.33525 moles Cl₂ ₓ (70.9g / mol) = 23.8g Cl₂

As yield of reaction is 74.7%, the real mass of Cl₂ you obtain (Actual yield) is:

23.8g Cl₂ ₓ 74% = 17.6g C₂

6 0

3 years ago

Pls answer this ASAP thank you

vaieri [72.5K]

Answer:

The anwer is not D the anwer is A

Explanation:

7 0

3 years ago