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4 years ago

You heat 3.869 g of a mixture of fe3o4 and feo to form 4.141 g fe2o3. What is the mass of oxygen reacted?

Chemistry

1 answer:

Lostsunrise [7]4 years ago

5 0

The reaction is

FeO + Fe3O4 + 1/2 O2---> 2Fe2O3

Thus as shown in the balanced equation two moles of Fe2O3 are formed when 0.5 moles of O2 reacted with mixture of FeO and Fe3O4

moles of Fe2O3 = MAss / Molar mass = 4.141 / 159.69 = 0.0259 moles

So moles of O2 needed = 0.5 X 0.0259 = 0.01295

Mass of O2 = moles X molar mass = 0.01295 X 32 = 0.4144 grams

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PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?

Leya [2.2K]

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

$n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}$

So, the value of n is 0.207 mol.

6 0

3 years ago

John has a solution with a pH of 12.21. what is the hydrogen ion concentration of this solution?

Georgia [21]

Answer:

In this way, ph is determined by hydrogen ion concentration . In the case of neutral solution [H+] =10-7 which we call a ph of 7. This means, for example, that a hydrogen-ion concentration of a solution with a ph of 4 is 10-4 mol/l, meaning it contains 0.0001 mole of hydrogen ion in a solution of 1 liter.

5 0

3 years ago

1 loop in the primary coil and 8 loops in the secondary. If the secondary voltage is 120 V, what must be the primary voltage

Jet001 [13]

Answer:

$V_{p} = 15 V$

Explanation:

<u>Given the following data;</u>

Number of loops in primary coil, Np = 1 loop.

Number of loops in secondary coil, Ns = 8 loops

Voltage in secondary coil, Vs = 120V

To find the voltage in the primary coil, Vp;

Transformer ratio is given by the formula;

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$

Making Vp the subject of formula;

$V_{p} = \frac{N_{p}}{N_{s}} * V_{s}$

Substituting into the equation, we have;

$V_{p} = \frac{1}{8} * 120$

$V_{p} = \frac{120}{8}$

$V_{p} = 15 V$

Therefore, the voltage in the primary coil, Vp is 15 Volts.

4 0

3 years ago

The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab

mixas84 [53]

Answer:

See the answers below

Explanation:

1) 100. mL of solution containing 19.5 g of NaCl (3.3M)

2) 100. mL of 3.00 M NaCl solution (3 M)

3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)

4) Number 1 and 5 have the same concentration (1.5M)

MW of NaCl = 23 + 36 = 59 g

For number 3

59 g ------------------- 1 mol

19,5 g ----------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For number 4,

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For number 5

Molarity (M) = 0.450/0.3 = 1.5 M

4 0

3 years ago

Read 2 more answers

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago