For the reaction #2HNO_3 + Mg(OH)_2 -> Mg(NO_3)_2 + 2H_2O#, how many grams of magnesium nitrate are produced from 8 moles of water?

Answer 1

#593.2 " g "Mg(NO_3)_2#

We can use stoichiometry to find the moles of #Mg(NO_3)_2# and then the respective grams of #Mg(NO_3)_2#.
If 8 moles of water are produced, we can find out how many moles of #Mg(NO_3)_2# are produced using this equation:
#(8 " "cancel("mol "H_2O))/1*(1 " mol " Mg(NO_3)_2)/(2" "cancel("mol "H_2O))=4" mol "Mg(NO_3)_2#
From here, we can use ratios to convert to the grams of #Mg(NO_3)_2#:
#(4 " "cancel("mol "Mg(NO_3)_2))/1*(148.3 " g " Mg(NO_3)_2)/(1 " "cancel("mol " Mg(NO_3)_2))=593.2 " g "Mg(NO_3)_2#
(You can get the molar mass of #Mg(NO_3)_2# by referencing a periodic table and summing the masses of each element in the compound.)
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Answer 2

The molar ratio between water and magnesium nitrate in the reaction is 2:1. Therefore, if 8 moles of water are consumed, 4 moles of magnesium nitrate will be produced. To find the mass of magnesium nitrate produced, you need to know its molar mass, which is approximately 148.3 g/mol. So, the mass of magnesium nitrate produced is:

4 moles × 148.3 g/mol = 593.2 grams

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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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