Question

A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical equation:
2NaOH + Fe(NO3)2 → NaNO3 + Fe(OH)2.

The reaction produced 3.70 grams of iron(II) hydroxide.

Assuming the reaction came to completion, what was the initial mass of sodium hydroxide?

Answer 1

Answer:

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Explanation:

The initial  mass  of sodium hydroxide  is  3.3 g (answer C)  calculation Step 1 : find the  moles of iron (ii) hydroxide (  Fe(OH)₂ moles =  mass÷  molar mass from periodic table the  molar mass of Fe(OH)₂  = 56 + [16 +1]2  = 90 g/mol moles  is therefore = 3.70 g÷ 90 g/mol = 0.041 moles Step 2:  use the mole ratio to  calculate the moles of  sodium hydroxide (NaOH)    from given equation  NaOH : Fe(OH)₂    is 2 :1 therefore the moles of NaOH = 0.041 x 2 = 0.082 moles Step 3: find  mass of NaOH mass = moles x molar mass from the periodic table the  molar mass of NaOH = 23 +16 +1  = 40 g/mol mass  = 0.082  moles x 40 g/mol = 3.3 g  

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