How many liters of H2 gas would you have if you have 2.52 moles of H2?

If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)

Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

$w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%$

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0

3 years ago

(Giving brainliest if u help mee)

Degger [83]

I think it would be B

7 0

3 years ago

Read 2 more answers

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas