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18 September, 22:37

Given 2 grams FeSO4 and excess NaOH, calculate how much Fe (OH) 2 can be produced?

Given 2 grams Fe (NO3) 3 and excess NaOH, calculate how much Fe (OH) 3 can be produced?

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  1. 19 September, 00:26
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    question 1

    The much of Fe (OH) 2 that can be produced is 1.188 grams

    calculation

    step 1: write the equation for reaction

    FeSO₄ + 2NaOH Fe (OH) ₂ + Na₂SO₄

    Step 2: calculate the moles of FeSO₄

    moles = mass/molar mass

    from periodic table the molar mass of FeSo₄ is 152 g/mol

    = 2 g/152 g/mol=0.0132 moles

    step 3:use the mole ratio of to determine the moles of Fe (OH) ₂

    from equation above FeSO4: Fe (OH) ₂ is 1:1 therefore the moles of Fe (OH) ₂ is also = 0.0132 moles

    step 4: find mass of Fe (OH) ₂

    mass = moles x molar mass

    From periodic table the molar mass of Fe (OH) 2 = 90 g/mol

    mass = 0.0132 moles x 90 g/mol = 1.188 grams of Fe (OH) ₂

    Question 2

    The much of Fe (OH) 3 that can be produced is 0.888 grams

    calculation

    Step 1: write the equation for reaction

    Fe (NO₃) ₃ + 3NaOH → Fe (OH) ₃ + 3NaNO₃

    Step 2: find the moles of fe (NO₃) ₃

    moles = mass/molar mass

    from periodic table the molar mass of Fe (NO₃) ₃ is = 242 g/mol

    moles = 2 g/242 g/mol = 0.0083 moles

    Step 3: use the mole ratio to determine the moles of Fe (OH) ₃

    from equation above Fe (NO₃) ₃: Fe (OH) ₃ is 1:1 therefore the moles of Fe (OH) ₃ is also = 0.0083 moles

    step 4: find mass of Fe (OH) ₃

    mass = moles x molar mass

    from periodic table the molar mass of Fe (OH) ₃ = 107 g/mol

    mass = 0.0083 moles x 107 g/mol = 0.888 grams
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