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ZanzabumX [31]
3 years ago
15

How many liters of hydrogen gas are formed from the complete reaction of 1.03 mol of C? Assume that the hydrogen gas is collecte

d at a pressure of 1.0 atm and temperature of 319 K . Express your answer using two significant figures.
Chemistry
1 answer:
aliina [53]3 years ago
8 0

Answer:

27 liters of hydrogen gas will be formed

Explanation:

Step 1: Data given

Number of moles C = 1.03 moles

Pressure H2 = 1.0 atm

Temperature = 319 K

Step 2: The balanced equation

C +H20 → CO + H2

Step 3: Calculate moles H2

For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2

For 1.03 moles C we'll have 1.03 moles H2

Step 4: Calculate volume H2

p*V = n*R*T

⇒with p = the pressure of the H2 gas = 1.0 atm

⇒with V = the volume of H2 gas = TO BE DETERMINED

⇒with n = the number of moles H2 gas = 1.03 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 319 K

V = (n*R*T)/p

V = (1.03 * 0.08206 *319) / 1

V = 27 L

27 liters of hydrogen gas will be formed

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bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
2 years ago
Calculate the pOH of a 0.0143 M NaOH solution at 25 ⁰C.
arsen [322]

Answer:

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Explanation:

You can find the pOH using the following equation:

pOH = -log[OH⁻]

Since NaOH dissociates into 1 Na⁺ and 1 OH⁻, the concentration of both ions is 0.0143 M.

pOH = -log[OH⁻]

pOH = -log[0.0143]

pOH = 1.845

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Answer:

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Calculate the number of moles CuCl2 necessary to make 50 mL of a 0.15M solution
tekilochka [14]
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n =  M * V

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3 years ago
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