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3 years ago

How many moles are in 4.24kg of Na2CO3?

1 answer:

7 0

Data:
4.24 Kg converting (grams) → 4240 g

Molar Mass of $Na_{2} CO_{3}$
Na = 2*23 = 46 amu
C = 1*12 = 12 amu
O = 3*16 = 48 amu
-------------------------
Molar Mass of $Na_{2} CO_{3}$ = 46 + 12 + 48 = 106 g/mol

If:

106 g → 1 mol
4240 g → y

Solving: Rule of three (directly proportional)
 $\frac{106}{4240} = \frac{1}{y}$
multiply cross
106*y = 4240*1
106y = 4240
$y = \frac{4240}{106}$
$\boxed{\boxed{y = 40\:moles}}\end{array}}\qquad\quad\checkmark$

Answer:
B. 40.0 moles

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How many moles of h2so4 will be produced from 8.21 moles of fes2?

Ierofanga [76]

Well, if the coefficients are both 1, then you will make 8.21 moles.

See the equation is:

x moles= (8.21mol FeS₂)×(1 mole H₂SO₄ ÷ 1 mole FeS₂) = 8.21 moles H₂SO₄

So if you change the coefficients of the equation, your product will be a different number.

Example:
x moles= (8.21mol FeS₂)×(2 mole H₂SO₄ ÷ 1 mole FeS₂) = 16.4 moles H₂SO₄

I hope this helps you out!
Brady

7 0

3 years ago

Read 2 more answers

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

Granite was formed slowly as magma cooled. What is the result of the slow cooling?

Aleksandr-060686 [28]

Igneous rocks from cooling magma. Granite is an igneous rock formed from magmathat cooled slowly underground. As the magma slowly cools, large mineral crystals form.

4 0

3 years ago

Read 2 more answers

The theoretical yield of NaBr from

Dvinal [7]

the percent yield of the reaction is 100%.

The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:

% yield = actual yield / theoretical yield * 100%

% yield of a reaction in this case Rate

In this case, the molar mass of NaBr is 102.9 g / mol, as you know:

444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g

theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g