If 507 g #FeCl_2# were used up in the reaction #FeCl_2 + 2NaOH -> Fe(OH)_2(s) + 2NaCl#, how many grams of #Fe(OH)_2# would be formed?

Answer 1

Simply apply some stoichiometry.

#507g# #FeCl_2# / #126.751 g# #FeCl_2# = #4 mol# #FeCl_2#
#4 mol# #FeCl_2# x #1 mol# #Fe(OH)_2# = #4 mol# #Fe(OH)_2#
#4 mol# #Fe(OH)_2# x #89.86 g# #Fe(OH)_2# = #359.44 g# #Fe(OH)_2#
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Answer 2

The molar mass of FeCl₂ is 126.75 g/mol. Therefore, 507 g of FeCl₂ is equivalent to 4 moles. According to the balanced equation, 1 mole of FeCl₂ produces 1 mole of Fe(OH)₂. So, 4 moles of FeCl₂ will produce 4 moles of Fe(OH)₂. The molar mass of Fe(OH)₂ is 89.86 g/mol. Hence, 4 moles of Fe(OH)₂ correspond to 359.44 g. Therefore, 359.44 grams of Fe(OH)₂ would be formed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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