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3 years ago

If 16 mol of oxygen were reacted with excess hydrogen gas, how many moles of water would be produced?

Chemistry

2 answers:

Elza [17]3 years ago

8 0

1) first, we need a balanced equation

2H₂ + O₂ ---> 2H₂O

2) we use the mole-to-mole ratio to convert the moles of oxygen to moles of water.

ratio ----> 1 mol O₂= 2 mol H₂O (based on the coefficients)

16 mol O₂= 2 x 16 mol H₂O
16 mol O₂ = 32 mol H₂O

jolli1 [7]3 years ago

6 0

I hope this helps...

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Why is water essential to living things? (1 pc

DIA [1.3K]

Answer:

Living organisms need water to survive. Many scientists even believe that if any extra-terrestrial exists, water must be present in their environments. All oxygen-dependent organisms need water to aid in the respiration process. Some organisms, such as fish, can only breathe in water. Other organisms require water to break down food molecules or generate energy during the respiration process. Water also helps many organisms regulate metabolism and dissolves compounds going into or out of the body.

Explanation:

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

Given 50g of sodium how many moles do you have?

Anastaziya [24]

Answer:

Amount of Na = 2.17moles

Explanation:

Mass of Na = 50g

Molar mass of Na = 23g/mol

Amount of mole = mass/molar mass

Amount of mole = 50/23

Amount of mole = 2.17moles

5 0

3 years ago

If the stack of bricks on the right has a mass of 240 kg, approximately what is the mass of the stack of bricks on the left?

solong [7]

Answer:

720

Explanation: study island

4 0

3 years ago

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$