Given the balanced equation :
3 CO + 7 H2 -> C3H8 + 3 H2O
How much C3H8 will be produced if 45.8 g CO reacts with 87.3 H2 ?
Please all your work
(i did it but im not sure if i did all the work right)
To determine the amount of C3H8 produced, we need to use the concept of stoichiometry, which relates the coefficients in a balanced chemical equation to the number of moles and masses of substances involved.
1. Calculate the number of moles of CO:
Given mass of CO = 45.8 g
Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol
Moles of CO = (45.8 g) / (28.01 g/mol) = 1.637 mol (rounded to three decimal places)
2. Calculate the number of moles of H2:
Given mass of H2 = 87.3 g
Molar mass of H2 = 1.008 g/mol
Moles of H2 = (87.3 g) / (2.016 g/mol) = 43.282 mol (rounded to three decimal places)
3. Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of CO and H2 using the coefficients in the balanced equation. The reactant that produces fewer moles of C3H8 will be the limiting reactant.
- For CO: 3 moles CO produces 1 mole C3H8
- For H2: 7 moles H2 produce 1 mole C3H8
Moles of C3H8 from CO = (1.637 mol CO) × (1 mol C3H8 / 3 mol CO) = 0.546 mol C3H8 (rounded to three decimal places)
Moles of C3H8 from H2 = (43.282 mol H2) × (1 mol C3H8 / 7 mol H2) = 6.183 mol C3H8 (rounded to three decimal places)
Since the moles of C3H8 from CO (0.546 mol) are less than those from H2 (6.183 mol), CO is the limiting reactant.
4. Calculate the moles of C3H8 produced from the limiting reactant:
From the balanced equation, 1 mole of CO produces 1 mole of C3H8.
Moles of C3H8 produced from the limiting reactant = 0.546 mol (rounded to three decimal places).
5. Calculate the mass of C3H8 using the moles and molar mass:
Molar mass of C3H8 = (3 × 12.01) + (8 × 1.008) = 44.10 g/mol
Mass of C3H8 produced = (0.546 mol) × (44.10 g/mol) = 24.065 g (rounded to three decimal places)
Therefore, approximately 24.065 g of C3H8 will be produced when 45.8 g of CO reacts with 87.3 g of H2.
To determine the amount of C3H8 produced, we need to use stoichiometry. Stoichiometry is a mathematical relationship between the amounts of reactants and products in a balanced chemical equation.
1. Start by calculating the molar mass of CO (carbon monoxide) and H2 (hydrogen gas):
- CO: C (1 atom) + O (1 atom) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
- H2: H (2 atoms) = 1.01 g/mol + 1.01 g/mol = 2.02 g/mol
2. Convert the given masses of CO and H2 to moles using the molar mass from step 1:
- Moles of CO = 45.8 g CO / 28.01 g/mol = 1.636 mol CO
- Moles of H2 = 87.3 g H2 / 2.02 g/mol = 43.168 mol H2
3. Use the coefficients from the balanced equation to determine the mole ratio between CO and C3H8. The coefficient in front of CO is 3, and the coefficient in front of C3H8 is also 3:
- Moles of C3H8 = 1.636 mol CO * (1 mol C3H8 / 3 mol CO) = 0.545 mol C3H8
4. Calculate the mass of C3H8 produced using the molar mass of C3H8:
- Mass of C3H8 = 0.545 mol C3H8 * 44.09 g/mol = 24.03 g C3H8
Therefore, 24.03 g of C3H8 will be produced when 45.8 g of CO reacts with 87.3 g of H2.