Ibuprofen can be found in 800 mg doses in over-the-counter analgesics, such as Advil and Motrin. How many grams of iburofen

Chemistry

1 answer:

geniusboy [140]2 years ago

7 0

0.8g Ibuprofen

1 g = 10^-3g = .001g

Ibuprofen has 800 mg doses in over-the-counter analgesic

800g = 800 × .001

= 0.8g

Ibuprofen is Nondteriodal Anti-inflammatory Drug (NSAID)
Ibuprofen's Mechanism of Action is Decreases inflammation, pain, and fever through inhibition of cyclooxygenase activity and prostaglandin synthesis
nonsteroidal anti-inflammatory medication (NSAID) used for pain relief and to reduce fever by stops inflammation and by blocking formation of cyclo-oxygenase (COX-2) a chemical mediator of inflammatory chemicals. i.e prostaglandins
It comes under the Class analgesic (reduce pain) and antipyretic (FIRE - reduce fever)
e side effects of ibuprofen NSAID are peripheral edema, fluid retention with edema, tinnitus, purpura, petechiae, anorexia, diarrhea, rash, nausea, vomiting, fatigue, dizziness, lightheadedness, anxiety, confusion

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$