3Ca(NO3)2(aq)+2Na3PO4(aq)->Ca3(PO4)2(s)+6NaNO3(aq)
A.)the moles Na3PO4 required to react with 2.03 L of 0.891 M Ca(NO3)2
B.)the grams of Ca3(PO4)2 that can be obtained from 136 mL of 0.627 M Ca(NO3)2
C.)the volume of 0.76 M Na3PO4 needed to react with 25 mL of 0.79 M Ca3(NO3)2
D.) the molarity (M) of the Ca(NO3)2 solution when 48.6 mL react with 52.8 mL of 4.1 M Na3PO4
What is it that you are having trouble with.
the whole problem im not sure what i have to do that's the reason why i posted this question on this website so someone can help me.
To solve these problems, we will use the concept of stoichiometry. Stoichiometry is a mathematical relationship between the reactants and products in a chemical reaction. It allows us to determine the quantities of substances involved in a reaction.
In this case, we are given the balanced chemical equation:
3Ca(NO3)2(aq) + 2Na3PO4(aq) -> Ca3(PO4)2(s) + 6NaNO3(aq)
Let's solve each problem step by step:
A.) The moles Na3PO4 required to react with 2.03 L of 0.891 M Ca(NO3)2:
We have the concentration (Molarity) and volume of Ca(NO3)2. To find the moles of Ca(NO3)2, we use the formula:
moles = concentration x volume
moles of Ca(NO3)2 = 0.891 M x 2.03 L
Now we need to calculate the moles of Na3PO4 using the stoichiometric ratio between Ca(NO3)2 and Na3PO4:
From the balanced equation, we see that the stoichiometric ratio between Ca(NO3)2 and Na3PO4 is 3:2. So, for every 3 moles of Ca(NO3)2, we have 2 moles of Na3PO4.
moles of Na3PO4 = (moles of Ca(NO3)2 / 3) x 2
Plug in the value we calculated for moles of Ca(NO3)2 and calculate moles of Na3PO4.
B.) The grams of Ca3(PO4)2 that can be obtained from 136 mL of 0.627 M Ca(NO3)2:
We have the concentration (Molarity) and volume of Ca(NO3)2. To find the moles of Ca(NO3)2, we use the formula:
moles = concentration x volume
moles of Ca(NO3)2 = 0.627 M x 0.136 L
Now we need to calculate the grams of Ca3(PO4)2 using the stoichiometric ratio between Ca(NO3)2 and Ca3(PO4)2:
From the balanced equation, we see that the stoichiometric ratio between Ca(NO3)2 and Ca3(PO4)2 is 1:1. So, for every 1 mole of Ca(NO3)2, we have 1 mole of Ca3(PO4)2.
moles of Ca3(PO4)2 = moles of Ca(NO3)2
Now we can calculate the grams of Ca3(PO4)2 using the molar mass of Ca3(PO4)2.
grams of Ca3(PO4)2 = moles of Ca3(PO4)2 x molar mass of Ca3(PO4)2
C.) The volume of 0.76 M Na3PO4 needed to react with 25 mL of 0.79 M Ca3(NO3)2:
We have the concentration (Molarity) and volume of Ca3(NO3)2. To find the moles of Ca3(NO3)2, we use the formula:
moles = concentration x volume
moles of Ca3(NO3)2 = 0.79 M x 0.025 L
Now we need to calculate the volume of Na3PO4 using the stoichiometric ratio between Ca3(NO3)2 and Na3PO4:
From the balanced equation, we see that the stoichiometric ratio between Ca3(NO3)2 and Na3PO4 is 2:3. So, for every 2 moles of Ca3(NO3)2, we have 3 moles of Na3PO4.
moles of Na3PO4 = (moles of Ca3(NO3)2 / 2) x 3
Plug in the value we calculated for moles of Ca3(NO3)2 and calculate moles of Na3PO4. Finally, convert moles of Na3PO4 to volume using the formula:
volume = moles / concentration
D.) The molarity (M) of the Ca(NO3)2 solution when 48.6 mL react with 52.8 mL of 4.1 M Na3PO4:
We have the volume and concentration of Na3PO4. To find the moles of Na3PO4, we use the formula:
moles = concentration x volume
moles of Na3PO4 = 4.1 M x 0.0528 L
Now we need to calculate the moles of Ca(NO3)2 using the stoichiometric ratio between Na3PO4 and Ca(NO3)2:
From the balanced equation, we see that the stoichiometric ratio between Na3PO4 and Ca(NO3)2 is 2:3. So, for every 2 moles of Na3PO4, we have 3 moles of Ca(NO3)2.
moles of Ca(NO3)2 = (moles of Na3PO4 / 2) x 3
Plug in the value we calculated for moles of Na3PO4 and calculate moles of Ca(NO3)2. Finally, calculate the molarity of Ca(NO3)2 using the formula:
Molarity = moles / volume
Plug in the value we calculated for moles of Ca(NO3)2 and the given volume, and calculate the molarity.