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3 years ago

- How many moles of H2O are required to react to form 2.5 grams of CH ?

Chemistry

1 answer:

abruzzese [7]3 years ago

7 0

Answer:

0.312 moles of H2O

Explanation:

no. of moles of ch4= mass ÷ molar mass

=2.5 ÷ 16.04

=0.156 moles of ch4

According to balanced chemical equation

CH4 : H2O

1 mole : 2 moles

0.156 moles : x moles

by cross multiplication

x= (0.156x2) ÷ 1

= 0.312 moles of H2O

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If a glass bottle breaks into thousands of tiny pieces what type of change occurs

mixas84 [53]

Physical change occurs.

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2 years ago

the mass by mass percentage of a solution is 20%. find the mass of the solute dissolved in 500g of solution

GuDViN [60]

Answer:

$m_{solute}=100g$

Explanation:

Hello,

In this case, by considering the mathematical definition of by mass percentage of a solute as shown below:

$\%m/m=\frac{m_{solute}}{m_{solution}}*100\%$

We are able to compute the mass of the solute in a 20% solution having 500 g of solution as follows:

$m_{solute}=\frac{m_{solution}*\%m/m}{100\%} =\frac{500g*20\%}{100\%}\\ \\m_{solute}=100g$

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6 0

3 years ago

When heated, KClO3 decomposes into KCl and O2. KClO3--> 2KCl+ 3O . If this reaction produced 62.6 g of KCl, how much O2 was p

Vladimir [108]

The solution to your problem is as follows:

2 KClO3 → 2 KCl + 3 O2

MW of KCL = 39.1 + 35.35 = 74.6 g/mole

62.6/74.6 = 0.839 moles KCl produced

0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced

Therefore, there are 40.27g of O2 produced during the reaction.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

5 0

3 years ago

#1: Which element has the same number of valence electrons as hydrogen (H)?

soldi70 [24.7K]

Valence electrons are those in the outer most orbital shell (the valence shell). Elements in the same column (group) have the same number of valence electrons. Hydrogen is in Group 1. If you were to look at a periodic table of the elements, Lithium is right under Hydrogen. You are correct, the answer is D. Lithium.

6 0

3 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago