How many grams of Ni(OH)2 are produced from the reaction of 45.0mL of 1.75 M NaOH solution?
Here is a stoichiometry problem worked out as an example. Just follow the steps. Remember moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of grams of Ni(OH)2 produced, we need to calculate the amount of Ni(OH)2 formed using stoichiometry and then convert it to grams.
First, let's write the balanced chemical equation for the reaction between NaOH and Ni(OH)2:
2 NaOH + NiCl2 -> 2 NaCl + Ni(OH)2
From the equation, we can see that 2 moles of NaOH react to produce 1 mole of Ni(OH)2.
Now let's calculate the number of moles of NaOH in the given volume:
moles of NaOH = concentration (M) x volume (L)
moles of NaOH = 1.75 M x 0.0450 L
moles of NaOH = 0.07875 moles
Since the stoichiometry of the reaction is 2 moles of NaOH : 1 mole of Ni(OH)2, we can determine the number of moles of Ni(OH)2 produced:
moles of Ni(OH)2 = (0.07875 moles NaOH) / (2 moles NaOH : 1 mole Ni(OH)2)
moles of Ni(OH)2 = 0.03938 moles
Finally, let's convert the moles of Ni(OH)2 to grams using the molar mass of Ni(OH)2. The molar mass of Ni(OH)2 can be calculated as:
molar mass of Ni(OH)2 = atomic mass of Ni + 2 * (atomic mass of O + atomic mass of H)
molar mass of Ni(OH)2 = 58.69 g/mol
grams of Ni(OH)2 = moles of Ni(OH)2 x molar mass of Ni(OH)2
grams of Ni(OH)2 = 0.03938 moles x 58.69 g/mol
grams of Ni(OH)2 = 2.31 grams
Therefore, 2.31 grams of Ni(OH)2 are produced from the reaction of 45.0 mL of 1.75 M NaOH solution.