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alekssr [168]
3 years ago
15

What is the oral dose of aluminum hydroxide for an adult being treated for hyperacidity? What is the oral dose of aluminum hydro

xide for an adult being treated for hyperacidity?
Chemistry
1 answer:
Vsevolod [243]3 years ago
8 0

Explanation:

Aluminum hydroxide is used as an antacid to counteract hyper acidity associated with the gastritis and peptic ulcer disease which includes duodenal ulcer and gastric ulcer and hiatal hernia.

<u> Oral dosage (oral suspension) </u>

Adults

Recommended OTC dose of aluminum hydroxide is 10 mL PO 5—6 times per day, after having meals and at bedtime.

Children

Recommended OTC dose of aluminum hydroxide is 5—15 mL PO every 3—6 hours, or 1 and 3 hours after having meals and at bedtime.

Infants

Recommended OTC dose of aluminum hydroxide is  1—2 mL/kg PO per dose and should be given 1—3 hours after having meals and at bedtime.

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Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(
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<u>Answer:</u> The theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CoCl_2.6H_2O = 4.00 g

Molar mass of CoCl_2.6H_2O = 238 g/mol

Putting values in equation 1, we get:

\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol

The chemical equation for the reaction of CoCl_2.6H_2O to form  [Co(NH_3)_4(H_2O)_2]Cl_2 follows:

CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O

By Stoichiometry of the reaction:

1 mole of CoCl_2.6H_2O produces 1 mole of [Co(NH_3)_4(H_2O)_2]Cl_2

So, 0.0168 moles of CoCl_2.6H_2O will produce = \frac{1}{1}\times 0.0168=0.0168mol of [Co(NH_3)_4(H_2O)_2]Cl_2

Now, calculating the mass of [Co(NH_3)_4(H_2O)_2]Cl_2 from equation 1, we get:

Molar mass of [Co(NH_3)_4(H_2O)_2]Cl_2 = 234 g/mol

Moles of [Co(NH_3)_4(H_2O)_2]Cl_2 = 0.0168 moles

Putting values in equation 1, we get:

0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g

To calculate the percentage yield of [Co(NH_3)_4(H_2O)_2]Cl_2, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 1.20 g

Theoretical yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 3.93 g

Putting values in above equation, we get:

\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%

Hence, the theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

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<span>On the periodic table, the majority of elements are classified as "Metals"

In short, Your Answer would be Option A

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