All categories

2 years ago

8CO + 17H2 → C8H18 + 8H2O

Chemistry

1 answer:

nekit [7.7K]2 years ago

4 0

$\frac{6.50mole CO}{1} \frac{17mole H2}{8mole CO}\frac{2 gram H2}{1mole H2} =27.625$

3 sig figs= 27.6

You might be interested in

In which condition does barium chloride conduct electricity? And why

Anuta_ua [19.1K]

Answer:

When barium chloride (BaCl 2) is dissolved in water, the water conducts electricity. In what form will the dissolved BaCl 2 be found? a. as Ba 2+ and Cl - ions b. as Ba atoms and Cl 2 molecules

Explanation:

3 0

1 year ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

3 years ago

Facts about heat absorption<br>and simple questions about heat absorption with answer

Mariulka [41]

<span>As heat energy reaches an object it can be absorbed in a similar manner to the way sponges absorb water. Heat enters an object, warming it. The longer the object is exposed to the heat source, the more heat it absorbs.</span>

8 0

3 years ago

A change of matter is a physical change <br>True or False

kodGreya [7K]

Answer:

true

Explanation:

hope it helps

8 0

2 years ago

Read 2 more answers

Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti

djverab [1.8K]

<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

= 80,000 g

Therefore;

$Moles of water = \frac{80,000g}{18.02 g/mol}$

= 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O available = 65.0 kg or 65,000 g

Molar mass Li₂O = 29.88 g/mol

Moles of Li₂O = 65,000 g ÷ 29.88 g/mol

= 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
This means, Li₂O is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles

= 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

= 4.0815 × 10⁴ g or

= 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0

2 years ago