How many grams of CO are needed with an excess of Fe2O3 to produce 35.0 g Fe
Solid sodium metal reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of of hydrogen are formed when 35 g of sodium are added to water?
first, write the balanced reactions, then convert everything from grams to moles, work out the right amount in moles, then convert back to grams or molecules.
To calculate the grams of CO needed with an excess of Fe2O3 to produce 35.0 g Fe, we need to follow a few steps:
1. Write a balanced chemical equation for the reaction:
Fe2O3 + 3CO -> 2Fe + 3CO2
2. Calculate the molar mass of Fe:
Molar mass of Fe = 55.85 g/mol
3. Convert the given mass of Fe (35.0 g) to moles:
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 35.0 g / 55.85 g/mol = 0.626 mol
4. From the balanced equation, the molar ratio between Fe and CO is 2:3. Therefore, the moles of CO required would be:
Moles of CO = (3/2) x Moles of Fe
Moles of CO = (3/2) x 0.626 mol = 0.939 mol
5. Finally, we can convert the moles of CO to grams:
Grams of CO = Moles of CO x Molar mass of CO
(Note: We need to determine the molar mass of CO to proceed with this step)
The molar mass of CO is calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of CO = Molar mass of C + Molar mass of O
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Now, we can determine the grams of CO:
Grams of CO = 0.939 mol x 28.01 g/mol = 26.37 g
Therefore, approximately 26.37 grams of CO are needed with an excess of Fe2O3 to produce 35.0 g Fe.
For the second question about the reaction of sodium with water, we can follow these steps:
1. Write the balanced chemical equation for the reaction:
2Na + 2H2O -> 2NaOH + H2
2. Calculate the molar mass of sodium (Na) and hydrogen (H2):
Molar mass of Na = 22.99 g/mol
Molar mass of H2 = 2.02 g/mol
3. Convert the given mass of sodium (35 g) to moles:
Moles of Na = Mass of Na / Molar mass of Na
Moles of Na = 35 g / 22.99 g/mol = 1.52 mol
4. From the balanced equation, the molar ratio between sodium (Na) and hydrogen (H2) is 2:1. Therefore, the moles of hydrogen formed would be half the moles of sodium:
Moles of H2 = (1/2) x Moles of Na
Moles of H2 = (1/2) x 1.52 mol = 0.76 mol
5. Finally, we can convert the moles of hydrogen to molecules:
Number of molecules of H2 = Moles of H2 x Avogadro's number
(Note: Avogadro's number is approximately 6.022 x 10^23)
Number of molecules of H2 = 0.76 mol x 6.022 x 10^23 molecules/mol ≈ 4.57 x 10^23 molecules
Therefore, approximately 4.57 x 10^23 molecules of hydrogen are formed when 35 g of sodium react with water.