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3 years ago

What mass of fes is formed if 9.42 g of fe reacts with 68.0 g of s8? 8fe(s) + s8(s) â†’ 8fes(s)?

1 answer:

4 0

From the equation, we see that the molar ratio of Fe : S required is:

8 : 1

The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265

The molar ratio is:

1 : 1.6

Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of

Fe : FeS is
1 : 1

So 0.168 moles of FeS will form. The mass of FeS will be:

Mass = 0.168 * (56 + 32)
Mass = 14.78 grams

14.78 grams of FeS will be formed.

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3 years ago

How many hydrogen atoms are in 89.5 g of<br> C6H6 ?<br> Answer in units of atoms.

Elodia [21]

Solution :

Molar mass of $C_6H_6$ is :

M = 6×12 + 6×1 g

M = 78 g

78 gram of $C_6H_6$ contains $6.022 \times 10^{23}$ molecules.

So, 89.5 gram of $C_6H_6$ contains :

$n = 6.022 \times 10^{23} \times \dfrac{89.5}{78}\\\\n = 6.91 \times 10^{23}$

Now, from the formula we can see that one molecule of $C_6H_6$ contains 2 hydrogen atom . So, number of hydrogen atom are :

$h = 2\times 6.91 \times 10^{23}\\\\h = 1.38 \times 10^{22}\ atoms$

Hence, this is the required solution.

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2 years ago

How to turn form moles to grams

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2 years ago

Read 2 more answers

What will the charge oxidiation state be for boron

artcher [175]

So, compounds of boron contain boron in a positive oxidation state, generally +3. The sum of oxidation numbers of all constituent atoms of a given molecule or ion is equal to zero or the charge of the ion, respectively. ... In most of the stable compounds of boron, its oxidation number is +3

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2 years ago

If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T

Veronika [31]

Answer:

$m_{MnO_2}=21.2gMnO_2$

Explanation:

Hello,

In this case, given the balanced reaction:

$2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O$

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

$m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-} *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2$

Best regards.

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3 years ago