Question

2 g of hydrogen reacting with 8 g of oxygen to give x g of water value of 'x' is​

Answer 1

Answer: The molecular masses of hydrogen, oxygen and water are 2 g/mole, 32 g/mole and 18 g/mole respectively. 2×2=4 g of hydrogen reacts with 1×32=32 g of oxygen to form 2×18=36 g of water. =9 g of water. 9 grams of water can be produced when 8 g of hydrogen reacts with 8 g oxygen.

Explanation:

Answer 2

Answer:

9g of $H_{2}O$ is produced.

Explanation:

The equation for the formation of water is-

$2H_{2} + O_{2}$ → $2H_{2}O$

Two moles of hydrogen molecules react with one mole of oxygen molecule to give two moles of water molecules.

In order to find the mass of water produced when 2g of hydrogen reacts with 8g of oxygen, we go through the following steps:

1. Find the moles of hydrogen and oxygen reacting by dividing mass with molar mass.

For $H_{2}$,

Molar mass- 2g/mol

Mass- 2g

moles= $\frac{2}{2} = 1$

For $O_{2}$,

Molar mass- 32g/mol

Mass- 8g

moles=$\frac{8}{32} = 0.25$

2. Find the limiting reagent by dividing the moles with coefficients.

For $H_{2}$,

$\frac{1}{2} = 0.5$

For $O_{2}$,

$\frac{0.25}{1} = 0.25$

The limiting reagent is $O_{2}$.

3. Find the moles of water obtained.

$O_{2}$       :      $H_{2}O$

1          :      2

0.25   :      $x$

$2 * 0.25 = 1 * x$

$0.5 = x$

4. Find the mass of water.

Molar mass of $H_{2}O$- 18mol/g

Moles of $H_{2}O$- 0.5mol

Mass of $H_{2}O$ = 18 × 0.5 = 9g