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Tatiana [17]
3 years ago
13

If you react 2.00 g of hydrogen completely using 15.87 g of oxygen to produce water, how much water (in grams) will you have?

Chemistry
1 answer:
leonid [27]3 years ago
8 0

Answer:

The amount (mass) of water we will have is 17.869 grams

Explanation:

The molar mass of hydrogen gas H₂ = 2.016 grams/mole

The molar mass of oxygen gas = 31.999 g/mol

Therefore, 2.00 g of hydrogen will give;

2.00/2.016 = 0.9921 moles of H₂ gas and

15.87 g of O₂ will give;

15.87/31.999 = 0.49595 moles

The reaction is as follows;

2H₂ (g) + O₂ (g) → 2H₂O (l)

Two moles of H₂ react with one mole of O₂ to produce two moles of H₂O

Therefore 0.9921 moles of H₂ will react with 0.9921/2 or 0.49595 moles of O₂ to produce 0.9921 moles of H₂O

From the above we note that all the H₂ and O₂ are completely consumed to form 0.9921 moles of H₂O

Molar mass of H₂O = 18.01528 g/mol

Number of moles = Mass/(Molar mass)

∴ Mass of H₂O = (Molar mass) × (Number of moles)

= 18.01528 g/mol × 0.9921 moles = 17.869 grams

Therefore the amount (mass) of water we will have = 17.869 grams.

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Greeley [361]

Answer:

Explanation:

Option B volume is the correct answer

5 0
3 years ago
A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?
Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

5 0
3 years ago
What is the empirical formula
Radda [10]

Answer:

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Explanation:

4 0
2 years ago
A teaspoon of powdered caffeine can be as deadly as a teaspoon of ricin. <br><br> True or False
Darina [25.2K]

A teaspoon of caffeine is <em>NOT</em> deadly as teaspoon of ricin                                                                                                            

4 0
3 years ago
Read 2 more answers
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
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