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Zarrin [17]
2 years ago
10

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

hydrogen and 20.0 g of iodine are heated, forming 10.0 g of hydrogen iodide, what mass of hydrogen remains unreacted? A. 10.0 g hydrogen remains B. 10.9 g hydrogen remains C. 15.0 g hydrogen remains D. 19.9 g hydrogen remains.
Chemistry
1 answer:
Kaylis [27]2 years ago
5 0

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

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IF your good at CHEMISTRY plls helpp!!!
gulaghasi [49]

molar concentration of Sn²⁺ ions = 0.273 M

molar concentration of Al³⁺ ions = 0.415 M

molar concentration of Cl⁻ ions = 1.692 M

Explanation:

First we calculate the number of moles of tin (II) chloride (SnCl₂) in the first solution:

number of moles = molar concentration × volume (L)

number of moles of SnCl₂ = 0.55 × 0.45 = 0.248 moles

And the number of moles of aluminum chloride (AlCl₃) in the second solution:

number of moles = molar concentration × volume (L)

number of moles of AlCl₃ = 0.7 × 0.65 = 0.455 moles

0.248 moles of SnCl₂ contains 0.248 moles of Sn²⁺ ions and 2 × 0.248 = 0.496 moles of Cl⁻ ions

0.455 moles of AlCl₃ contains 0.455 moles of Al³⁺ ions and 3 × 0.455 = 1.365 moles of Cl⁻ ions

Volume of the final solution = 450 mL + 650 mL = 1100 mL = 1.1 L

molar concentration = number of moles / volume (L)

molar concentration of Sn²⁺ ions = 0.248 / 1.1 = 0.273 M

molar concentration of Al³⁺ ions = 0.455 / 1.1 = 0.415 M

molar concentration of Cl⁻ ions = (0.496 + 1.365) / 1.1 = 1.692 M

Learn more about:

molar concentration

brainly.com/question/14198342

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#learnwithBrainly

6 0
3 years ago
A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.
Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

HBr+KOH\rightarrow KBr+H_2O

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

pOH=-log([OH^-])=-log(0.320)=0.50

Thus, the pH is:

pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

n_{KOH}^{remaining}=0.00425mol

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M

So the pOH and the pH turn out:

pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15

Best regards!

7 0
3 years ago
What happens if 2 hydrogen and 1 oxygen atoms are combined?
Wittaler [7]

Hello There!

We will have water (H2O)

So if you drink a glass of water, you drink a glass of H2O.

Hope it helps! :)

GraceRosalia here to help you :)

8 0
2 years ago
Read 2 more answers
a scientist is mixing a chemical for a solution foe an experiment. The solution contains 5/8 ounce of water and 1/6 ounce of sal
chubhunter [2.5K]
One whole and two eigths
6 0
3 years ago
Listed below are sets of elements.
Anna11 [10]
The correct answer is B) Chlorine, Sulfur, and Silicon

I'm 100% sure this is correct

Brainliest please!!!
7 0
3 years ago
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