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frutty [35]
3 years ago
8

Magnesium and Hydrochloric acid (HCl) react to form magnesium chloride (MgCl2) and hydrogen gas. If you started with 65.8 g of m

agnesium and 45.7 g of hydrochloric acid. What is the limiting reactant? Ensure you post calculations to prove your answer. (3 marks)
Chemistry
1 answer:
DiKsa [7]3 years ago
4 0

Answer:

Cl

Explanation:

n for Mg=m:Mm

=65.8:24

=2.74mol

n for Cl=m:Mm

=45.7:70

=0.65mol

According to the <em>number of moles</em> we can say that the limiting reagent is Cl as it has few moles.

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2.
katovenus [111]

Answer:

A. 6atm

Explanation:

Using pressure law equation:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

According to this question;

P1 = 3 atm

P2 = ?

T1 = 120K

T2 = 240K

Using P1/T1 = P2/T2

3/120 = P2/240

Cross multiply

240 × 3 = P2 × 120

720 = 120P2

P2 = 720/120

P2 = 6atm

4 0
2 years ago
What is four thing the atmosphere does for us
artcher [175]
Keeps in heat, protects us from radiation, provides oxygen, and protects us from objects coming towards the earth.
4 0
3 years ago
Read 2 more answers
Calculate the mass of butane needed to produce 97.4 g of carbon dioxide. Express your answer to three significant figures and in
Vsevolod [243]

Answer:

32.1 g

Explanation:

Step 1: Write the balanced combustion reaction

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 97.4 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

97.4 g × 1 mol/44.01 g = 2.21 mol

Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide

The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol

Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

0.553 mol × 58.12 g/mol = 32.1 g

4 0
3 years ago
What does evaporation produce?
baherus [9]

Answer:

C. Water Vapor

Explanation:

Hope this helps! Have an Amazing day!!

7 0
2 years ago
Read 2 more answers
To prepare 250.0 mL of 2.5 M KCl you will need to dilute _____ mL of 8.0 M KCl solution to a volume of 250.0 mL.
Radda [10]

Answer:

78.125ml

Explanation:

Number of moles in 250ml of 2.5M KCl is (250÷1000)litres×2.5M so we divide these moles by 8M. The answer gotten will be in litres so multiply by 1000 to get it in ml

6 0
3 years ago
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