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Salsk061 [2.6K]
3 years ago
15

If you have 6.25g of O2, how many moles of H2O are produced?

Chemistry
1 answer:
denis23 [38]3 years ago
8 0

Answer:

0.4 moles of water produced by 6.25 g of oxygen.

Explanation:

Given data:

Mass of oxygen = 6.25 g

Moles of water produced = ?

Solution:

Chemical equation;

2H₂ + O₂  → 2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 6.35 g/ 32 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of oxygen with water:

                    O₂          :          H₂O

                     1            :            2

                   0.2          :        2×0.2 = 0.4 mol

0.4 moles of water produced by 6.25 g of oxygen.

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Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you
Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

         = 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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2 years ago
To balance the reaction what coefficients (numbers) are needed: HBr +KOH ---&gt; KBr + H2O
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5 0
3 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

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What do you need to adjust to balance a chemical equation
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