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DiKsa [7]
3 years ago
11

If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

Chemistry
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

              2H₂  + O₂  -----------> 2H₂O

Now look at the reaction for mole ratio

             2H₂  + O₂  -----------> 2H₂O

             2 mole                       2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂  = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

                  2H₂         +       O₂        ----------->    2H₂O

          2 mole (2 g/mol)                                 2 mole (18 g/mol)

                    4 g                                                     36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

                       4 g of H₂ ≅ 36 g of H₂O

                        30 g of H₂ ≅ X of H₂O

Do cross multiplication

                  X of H₂O =  30 g x 36 g / 4 g

                  X of H₂O =  270 g

Now convert grams of H₂O into moles

               No. of moles = mass in grams/molar mass

Put values in above formula

               No. of moles = 270 g / 18 (g/mol)

               No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

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Which kind of matter can be homogenous or heterogenous ?
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Hope this helps you!
5 0
3 years ago
PLEASE SOMEONE HELP ME WITH THIS!
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Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

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3 years ago
How many molecules of ammonia are formed in the following equation: N2+3H2→2NH3
Katena32 [7]

Answer:

  • <u>two molecules of ammonia are formed by the reaction of one nitrogen and three hydrogen molecules.</u>

Explanation:

The  balanced chemical equation provides information on:

  • <u>Reactants</u>: those are the compounds that appear of the left side of the equation, each with its chemical formula.

  • <u>Products</u>: those are the compounds that appear on the right side of the equation, again, each with its chemical formula.

  • <u>Ratio</u>: the coefficients of each compound (the number to the left of the chemical formula) represent the ratio of the number of molecules that react and are formed.

In the given equation you have:

  • Equation: N₂ + 3H₂ → 2NH₃

  • The coefficients are 1 for nitrogen, 3 for hydrogen, and 2 for ammonia. Hence, 2 molecules of ammonia are formed by the reaction of 1 molecule of nitrogen and 3 molecules of hydrogen.
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Convert 1.45 × 1024 atoms of carbon to moles of carbon.
anygoal [31]
1 mol = 6.02 * 10^23 atoms of carbon
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1/x =6.02*10^23 / 1.45 * 10^24

6.02 * 10^23 x = 1.45 * 10^24
x = 1.45 * 10^24 / 6.02 * 10^23
x = 2.41 mols of carbon
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