How many grams of iron(III) nitrate.9H2O could be formed from the reaction of 3.676 g of iron with excess HNO3?
Please can u explain? i try balance the question i couldnt.
2Fe + 6HNO3 + 9H2O ==>2Fe(NO3)3.9H2O + 3H2
1. Write and balance the equation as above.
2. Convert 3.676 g Fe to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Fe to moles of the product.
4. Now convert moles of the product to grams. g = moles x molar mass.
Check my work. Post your work if get stuck.
the answer is 11.17 or not
To answer this question, you first need to balance the chemical equation for the reaction between iron and HNO3. The balanced equation is:
2 Fe + 6 HNO3 → 2 Fe(NO3)3 + 3 H2O
From the balanced equation, you can see that the reaction produces two moles of Fe(NO3)3 for every two moles of Fe used.
Next, you need to calculate the molar mass of Fe(NO3)3·9H2O (iron(III) nitrate.9H2O) to convert from moles to grams:
Fe(NO3)3·9H2O = (55.85 g/mol + 3 * (14.01 g/mol + 3 * 16.00 g/mol)) + (9 * 2.02 g/mol)
Fe(NO3)3·9H2O = 404.00 g/mol
Now, we can calculate the number of moles of Fe(NO3)3·9H2O that can be formed from 3.676 g of iron:
moles of Fe = mass / molar mass
moles of Fe = 3.676 g / 55.85 g/mol ≈ 0.0658 mol of Fe
Since the reaction produces two moles of Fe(NO3)3 for every two moles of Fe used, the number of moles of Fe(NO3)3·9H2O formed will also be 0.0658 mol.
Finally, you can convert the moles of Fe(NO3)3·9H2O to grams:
mass = moles * molar mass
mass = 0.0658 mol * 404.00 g/mol ≈ 26.61 g
Therefore, approximately 26.61 grams of iron(III) nitrate.9H2O can be formed from the reaction between 3.676 grams of iron and excess HNO3.