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3 years ago

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When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2 AlCl 3 ( aq )

+ 3 H 2 ( g ) What volume of H 2 ( g ) is produced when 3.60 g Al ( s ) reacts at STP?

1 answer:

Dovator [93]3 years ago

4 0

Answer:

The volume of hydrogen gas produced at STP is 4.90 liters.

Explanation:

$2Al (s) + 6 HCl (aq)\rightarrow 2 AlCl_3 (aq) + 3 H_2 (g)$

Moles of aluminium = $\frac{3.60 g}{27 g/mol}=0.1333 mol$

According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.

Then 0.1333 moles of aluminium will give:

$\frac{3}{2}\times 0.1333 mol=0.2 moles$ of hydrogen gas

Volume of 0.2 moles of hydrogen gas at STP = V

Temperature at STP = T = 298.15 K

Pressure at STP = P = 1 atm

n = 0.2 mol

PV = nRT (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.2mol\times 0.0821 atm L/mol K\times 298.15 K}{1 atm}=4.8956 L\approx 4.90 L$

The volume of hydrogen gas produced at STP is 4.90 liters.

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A mixture of 50.0 mL of ammonia gas and 60.0 mL of oxygen gas react. If all the gases are at the same temperature and pressure,

fredd [130]

Answer:

72.0 mL of steam is formed.

Explanation:

The reaction is :

$4 NH_{3} + 5O_{2} \rightarrow 4NO+6H_{2} O$

You can treat coefficient of compounds as amount of volume used.

Therefore for 4 mL of ammonia 5 mL of oxygen is used to form 4 mL of nitric oxide gas and 6 mL of steam.

For 1 mL of ammonia $\frac{5}{4}$ (=1.25) mL of oxygen is used to form $\frac{4}{4}$ (=1) mL of nitric oxide gas and $\frac{6}{4}$ (=1.5) mL of steam.

OR

Just transform the chemical equation by dividing the whole equation by 4 so that the coefficient of $NH_{3}$ become one like this

$NH_{3} + \frac{5}{4} O_{2} \rightarrow \frac{4}{4}NO+\frac{6}{4}H_{2} O$

We don't know which one will be completely exhausted and which one will be left so we have to consider two cases :

<em>1. </em><em>Assume ammonia to be completely exhausted</em>

For 50 mL of ammonia $50 \times \frac{5}{4}$ (= 62.5) mL of oxygen is needed. But we have just 60 mL of oxygen so this assumption is false.

2. <em>Assume oxygen to be completely exhausted</em>

For 60 mL of oxygen only $60 \times \frac{4}{5}$ (=48) mL of ammonia is needed. In this case we have sufficient amount of ammonia. So this case is true.

$60\times\frac{4}{5}NH_{3} + 60\ O_{2} \rightarrow 60\times\frac{4}{5}NO+60\times\frac{6}{5}H_{2} O\\\\48NH_{3} + 60\ O_{2} \rightarrow 48NO+72H_{2} O$

Now we know that during complete reaction 48 mL of ammonia and 60 mL of oxygen is used which will form $60 \times \frac{4}{5}$ (= 48) mL of nitic oxide gas and $60 \times \frac{6}{5}$ (= 72) mL of steam.

Therefore <em>72 mL of steam </em>is formed.

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3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago