Help with this chemistry question please???

Chemistry

1 answer:

givi [52]3 years ago

6 0

To find percent composition, you should first find the total of all of the components. Just add them all together:

$100g+200g+50g=350g$

Then, divide the amount of the component you are interested in, in this case, chocolate (50g), by the total amount to give:

$\frac{50g}{350g}$

This will give you the answer in decimal form, but to convert to a percent, you need to multiply by 100:

$\frac{50g}{350g}*100$

Therefore, the answer choice that is correct is A.

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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 $AgC_{10}H_9N_4SO_2$