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3 years ago

Please help me ill give brainiest

Chemistry

2 answers:

V125BC [204]3 years ago

6 0

Reaction on the left product on the right

lyudmila [28]3 years ago

5 0

Your reactants are on the left side and your product is on the right side

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Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2

arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ : 159.69 g/mol

mol Fe₂O₃

$\tt \dfrac{100}{159,69}=0.626$

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

$\tt \dfrac{2}{1}\times 0.626=1.252$

mass of Fe(Ar=55.845 g/mol) :

$\tt 1.252\times 55.845=69.92~g$

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

$\tt =\dfrac{65}{69.92}=0.93=93\%$

8 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago

Which has the greater concentration of hydrogen ions, a substance with a pH of 5 or a substance with a pH of 4?

evablogger [386]

Answer:

pH 4

Explanation:

Firstly, we define pH as the negative logarithm to base 10 of the concentration of hydrogen ions.

Mathematically, we express this as:

pH = -log[H+]

Now let’s us calculate the concentration of hydrogen in each of the pH

For pH 4, we have:

4 = -log[H+]

[H+] = -Antilog(4)

[H+] = 0.0001M

For pH 5,

[H+] = -Antilog(5)

[H+] = 0.00001M

We can see that 0.0001 is greater than 0.00001 and thus it has a greater concentration of hydrogen ions

4 0

4 years ago

What would be the volume of 0.500 mole of an<br> ideal gas at STP

Katena32 [7]

I think v=11.21 right?

8 0

3 years ago

How many kilocalories are required to increase the temperature of 15.6 g of iron from 122 °c to 355 °c. the specific heat of iro

Dmitriy789 [7]

Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,

$Q = m s dT$

Where, m is mass of iron

s is specific heat of iron

d T is change in temperature in celcius

$= 15.6 g * 0.45 J /g /C * (355 - 122) = 1.63 * 10^3 J$

1 cal = 4.2 J

Then,

$Q = (1.63 * 10^3) /4.2 = 0.389 K cal$

Thus 0.389 k cal of enrgy is required by a 15.6 g Fe to reach to 355 C^0

4 0

3 years ago