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Liono4ka [1.6K]

3 years ago

15

Aluminum metal and manganese (IV) oxide react to produce aluminum oxide and manganese metal. What mass of manganese (IV) oxide r

eacts to produce 1.12kg of manganese metal?

1 answer:

gayaneshka [121]3 years ago

6 0

Answer: 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

Explanation:

The balanced chemical equation is:

$4Al+3MnO_2\rightarrow 2Al_2O_3+3Mn$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of manganese = $\frac{1.12\times 1000g}{55g/mol}=20.4mol$

According to stoichiometry :

3 moles of $Mn$ is produced by = 3 moles of $MnO_2$

Thus 20.4 moles of $Mn$ is produced by = $\frac{3}{3}\times 20.4=20.4moles$ of $MnO_2$

Mass of $MnO_2=moles\times {\text {Molar mass}}=20.4moles\times 86.9g/mol=1773g=1.773kg$ (1kg=1000g)

Thus 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

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Answer:

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An element X forms a compound XH 3 with hydrogen. Another element Y forms a compound YX 2 with X. Given that the valency of hydr

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X $H_{3}$
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3 years ago

Fill in the blank:

Roman55 [17]

Answer:

Weathering, Erosion

Explanation:

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Hope this helps

All the love, Ya boi Fraser :)

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2 years ago

Read 2 more answers

A plate moves 200 m in 10,000 years. What is its rate in cm/year?

Varvara68 [4.7K]

Answer:

The answer is

<h2>2 cm/year</h2>

Explanation:

To find the rate in cm/year we must first convert 200 m into cm

1 m = 100 cm

if 1 m = 100 cm

Then 200 m = 200 × 100 = 20 ,000 cm

So the rate is

<h2> $\frac{20000}{10000}$ </h2>

<u>Reduce the fraction with 10,000</u>

We have the final answer as

<h3>2 cm/year</h3>

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2 years ago

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

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$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

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$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

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2 years ago