How is stoicheometry used to keep camels alive

According to the following reaction, how many grams of water are produced in the complete reaction of grams of sulfuric acid?

Natasha2012 [34]

This question requires the mass of sulfuric acid and a balance equation. The complete question is given below

Question:

According to the following reaction, how many grams of water are produced in the complete reaction of 24.1 grams of sulfuric acid?

H₂SO₄(aq) + Zn(OH)₂(s) ---------> ZnSO₄(aq) + 2H₂O(l)

Answer:

8.85 grams of water is obtained from 24.1 grams of sulfuric acid according to the given reaction

Explanation:

In this problem, the mass of water can be determined by using the balanced chemical equation.

Step 1: Write all data

Molar mass of water = 18 g

Molar mass of sulfuric acid = 98 g

Given mass of sulfuric acid = 24.1 g

Mass of water from reaction = ?

Step 2: Write statement for conversion

Given equation shows that

1 mole of H₂SO₄ gives 2 moles of water

Step 3: Convert moles into molar mass

Convert the moles into molar mass so, the statement becomes,

98 g of H₂SO₄ gives (2)(18) g of water

1 g of H₂SO₄ gives (2)(18)/98 g of water

Step 4: Use given data

24.1 g of H₂SO₄ gives (2)(18)(24.1)/98 g of water

amount of water = 8.85 g

5 0

2 years ago

Plsss helpppp im confused will mark BRAINLIEST

mr_godi [17]

I am thinking Li & CI :)

6 0

3 years ago

Read 2 more answers

What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as

serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

CaCl2 ---> Ca2+ + 2 Cl-

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

Cl- = 0.4 moles (ANSWER)

6 0

3 years ago

Help me ASAP PLS

n200080 [17]

Answer:

18 g

Explanation:

We'll begin by converting 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 0.2 M

Mole of C₆H₁₂O₆ =?

Molarity = mole / Volume

0.2 = Mole of C₆H₁₂O₆ / 0.5

Cross multiply

Mole of C₆H₁₂O₆ = 0.2 × 0.5

Mole of C₆H₁₂O₆ = 0.1 mole

Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.1 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mass = mole × molar mass

Mass of C₆H₁₂O₆ = 0.1 × 180

Mass of C₆H₁₂O₆ = 18 g

Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.

6 0

2 years ago

What is the mass of 0.0150 moles of Na₂SO4 (molar mass = 142.04 g/mol)?

olga2289 [7]

Answer:

2.13 g

Explanation:

(142.04)(0.0150) = 2.13 g

4 0

1 year ago

How is stoicheometry used to keep camels alive​

How is stoicheometry used to keep camels alive