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m_a_m_a [10]
2 years ago
11

3. What is the greatest amount of H2O that can be made with 3.8 moles of H and 5 moles of

Chemistry
1 answer:
Pachacha [2.7K]2 years ago
3 0

Answer:  Hello!

first i believe we need a balanced equation to start...

i got 2H2 + 1O2 = 2H2O

This tells us that we need 2 moles of H2 for every 1 mole of O2 Since we only have 1 mole of H2 compared to the 5 moles of O2 hydrogen is the limiting reagent. For illustration, divide the balanced equation by 2 in order to get 1 mole of H2 If we start with 1.0 moles of H2 we'll produce 1.0 mole of H2O

Your welcome <3

Explanation:

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To help spread seeds to make more strawberries. Once a person bites into a strawberry some of the seeds fall and help plant new ones.

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7 0
4 years ago
Đơn chất nào sau đây tác dụng với dung dịch axit sunfuric loãng sinh ra khí ?
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3 years ago
A student conducts an experiment by placing a wooden stick in a cup of sugar solution. Over the next few weeks, sugar crystals f
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8 0
3 years ago
C(S)+O2(g)--&gt;CO2(g)
soldi70 [24.7K]

<u>Answer:</u> The correct answer is 1.18 g.

<u>Explanation:</u>

We are given a chemical equation:

C(S)+O2(g)\rightarrow CO_2(g)

We know that at STP conditions:

22.4L of volume is occupied by 1 mole of a gas.

So, 2.21L of carbon dioxide is occupied by = \frac{1}{22.4L}\times 2.21L=0.0986mol of carbon dioxide gas.

By Stoichiometry of the above reaction:

1 mole of carbon dioxide gas is produced by 1 mole of carbon

So, 0.0986 moles of carbon dioxide is produced by = \frac{1}{1}\times 0.0986=0.0986mol of carbon.

Now, to calculate the mass of carbon, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of carbon = 0.0986 mol

Molar mass of carbon = 12 g/mol

Putting values in above equation, we get:

0.0986mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=1.18g

Hence, the correct answer is 1.18 g.

3 0
3 years ago
Read 2 more answers
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Answer:

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hydrogen peroxide - water + oxygen ( single displacement reaction)

Hope it helps :)

8 0
3 years ago
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