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Nadya [2.5K]
3 years ago
14

Co(g) + 2 h2 --> ch3oh 2.50 g of hydrogen is reacted with 30.0 l of carbon monoxide at stp. 1. what is the limiting reactant?

*hint: only list the element symbol* 2. what mass of ch3oh is produced? *hint: only list the grams* 3. how much excess is left over? *hint: only list the grams*
Chemistry
2 answers:
Lemur [1.5K]3 years ago
7 0
CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)

2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover

anastassius [24]3 years ago
7 0

Answer 1 : The limiting reactant is, hydrogen gas, H_2

Solution :

First we have to calculate the moles of hydrogen gas.

\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{2.5g}{2g/mole}=1.25moles

Now we have to calculate the volume of hydrogen gas.

As, 1 mole of gas contains 22.4 L volume of gas

So, 1.25 mole of hydrogen gas contains 1.25\times 22.4=28L volume of hydrogen gas

Now we have to calculate the limiting and excess reactant.

The given balanced reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(l)

From the balanced reaction, we conclude that

As, 44.8 L of hydrogen gas react with 22.4 L of carbon monoxide gas

So, 28 L of hydrogen gas react with \frac{22.4L}{44.8L}\times 28L=14L of carbon monoxide gas

The excess carbon monoxide = 30 L - 14 L = 16 L

Thus, the carbon monoxide is an excess reactant because it is present in excess amount and hydrogen gas is a limiting reactant because it is present in limited amount.

Answer 2 : The mass of CH_3OH is, 20 grams

Solution :

From the balance reaction, we conclude that

2 moles of hydrogen gas react to give 1 mole of CH_3OH

1.25 moles of hydrogen gas react to give \frac{1.25}{2}=0.625 mole of CH_3OH

Now we have to calculate the mass of CH_3OH

\text{Mass of }CH_3OH=\text{Moles of }CH_3OH\times \text{Molar mass of }CH_3OH

\text{Mass of }CH_3OH=(0.625mole)\times (32g/mole)=20g

The mass of CH_3OH is, 20 grams

Answer 3 : The amount of excess reactant is, 20 grams

The excess reactant is carbon monoxide.

As, 22.4 L volume of carbon monoxide gas has 28 gram of carbon monoxide gas

So, 16 L volume of carbon monoxide gas has \frac{28}{22.4}\times 16=20 gram of carbon monoxide gas

Thus, the The amount of excess reactant is, 20 grams

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Nastasia [14]
<h3>Answer:</h3>

0.918 M

<h3>Explanation:</h3>

Assuming the question requires we calculate the Molarity of sulfuric acid:

We are given:

  • Volume of the acid, H₂SO₄ = 25.00 ml
  • Volume of the base, KOH = 27.00 mL
  • Molarity of the base, KOH is 1.70 M

We can calculate the molarity of the acid using the following steps;

<h3>Step 1: Write the chemical equation for the reaction.</h3>

The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.

Therefore, the balanced equation will be;

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

<h3>Step 2: Determine the moles of the base, KOH </h3>

When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.

Number of moles = Molarity × Volume

                             = 1.700 M × 0.027 L

                              = 0.0459 moles

Thus, moles of KOH used is 0.0459 moles

<h3>Step 3: Determine the number of moles of the Acid, H₂SO₄</h3>

From the reaction, 1 mole of the acid reacts with 2 moles of KOH

Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2

Thus, moles of H₂SO₄ = Moles of KOH ÷ 2

                                     = 0.0459 moles ÷ 2

                                     = 0.02295 moles

<h3>Step 4: Calculate the molarity of the Acid </h3>

Molarity is the concentration of a solution in moles per liter

Molarity = Moles ÷ Volume

Molarity of the acid = 0.02295 moles ÷ 0.025 L

                                = 0.918 M

Thus, the molarity of the acid, H₂SO₄ is 0.918 M

5 0
3 years ago
An aqueous solution contains 0.23 M potassium hypochlorite.
Arturiano [62]

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO  </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em>  will produce NH₄⁺. 0.25mol HCl consume all NH₃.

5 0
3 years ago
A solution is prepared by dissolving 0.23 mol of chloroacetic acid and 0.27 mol of sodium chloroacetate in water sufficient to y
Sergio [31]

Answer:

c. chloroacetate ion

Explanation:

The chloroacetic acid, ClCH₂CO₂H, is a weak acid with Ka = 1.36x10⁻³. When this weak acid is in solution with its conjugate base, ClCH₂CO₂⁻ (From sodium chloroacetate) a buffer is produced. The addition of a strong acid as the HCl produce the following reaction

HCl + ClCH₂CO₂⁻ → ClCH₂CO₂H + Cl⁻.

Where the acid reacts with the chloroacetate ion to produce more chloroacetic acid

That means, the HCl reacts with the chloroacetate ion present in the buffer solution

Right answer is:

<h3>c. chloroacetate ion</h3>
8 0
2 years ago
A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A
Gemiola [76]

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!

5 0
3 years ago
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