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2 years ago

Given the reaction below, if 0.003 g of carbon dioxide is used up, how many grams of oxygen will be produced? (show work)

1 answer:

6 0

Stoichiometry: 6 CO2 (reactant) = 6 O2 (product)
Every 6 CO2 (atoms, moles, etc…) that react will wield 6 O2 (atoms, moles, etc…)
0.003g CO2 * (6 O2 / 6 CO2) = 0.003g O2

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In the following reaction, how many grams of silver can be produced from 49.1 g copper?

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166.4 g Ag grams of silver can be produced from 49.1 g of copper.

A mole is a very important unit of measurement that chemists use. A mole of something means you have 602,214,076,000,000,000,000,000 of that thing, like how having a dozen eggs means you have twelve eggs.

$1 mol Cu + 2 mol AgNO_3$ → $2 mol Ag + 1 mol Cu(NO3)2$

63.55 g Cu —> 2 x 107.688 g Ag

63.55 g Cu gives 215.376 g of Ag

So, 49.1 g Cu —> $\frac{215.376 g X 49.1 g}{63.55 g}$

= 166.4 g Ag

Hence, 166.4 g Ag grams of silver can be produced from 49.1 g of copper.

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2 years ago

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

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Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$

From the reaction, we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 2 mole of $AgC_{10}H_9N_4SO_2$

So, 0.1998 mole of $C_{10}H_{10}N_4SO_2$ react with 0.1998 mole of $AgC_{10}H_9N_4SO_2$

Now we have to calculate the mass of $AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g$

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0

3 years ago