Question

How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Answer 1

Answer:

4.85 grams of H2 will be produced

Explanation:

Step 1: Data given

Mass of HCl = 175 grams

Molar mass of HCl = 36.46 g/mol

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

2HCI + Na —> 2NaCl +H2

Step 3: Calculate moles HCl

Moles HCl = mass HCl / molar mass HCl

Moles HCl = 175 grams / 36.46 g/mol

Moles HCl = 4.80 moles

Step 4: Calculate moles H2

For 2 moles HCl we need 1 mol Na to produce 2 moles NaCl and 1 mol H2

For 4.80 moles HCl we'll have 4.80/ 2 = 2.40 moles H2

Step 5: Calculate mass H2

Mass H2 = moles H2 * molar mass H2

Mass H2 = 2.40 moles * 2.02 g/mol

Mass H2 = 4.85 grams

4.85 grams of H2 will be produced

Answer 2

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

• HCl: 2 moles
• Na: 1 mole
• NaCl: 2 moles
• H₂: 1 mole

You know the following masses of each element:

• H: 1 g/mole
• Cl: 35.45 g/mole
• Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

• HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
• Na: 23 g/mole
• NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
• H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

• HCl: 2 moles* 36.45 g/mole= 72.9 g
• Na: 1 mole* 23 g/mole= 23 g
• NaCl: 2 moles* 58.45 g/mole= 116.9 g
• H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

$mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}$

mass of H₂= 4.8 g

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium