Use the following balanced equation.
3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2
How many mL of 0.10 M Sr(OH)2 are required to react with 30 mL of 0.30 M H3PO4?
moles H3PO4 = M x L = ?
Using the coefficients in the balanced equation, convert moles H3PO4 to moles Sr(OH)2.
Now M Sr(OH)2 = moles Sr(OH)2/L Sr(OH)2
You have M and moles, solve for L and convert to mL.
To determine the volume of 0.10 M Sr(OH)2 required, we need to use the stoichiometry of the balanced equation and the given concentrations and volumes of the reactants.
The balanced equation tells us that the stoichiometric ratio between Sr(OH)2 and H3PO4 is 3:2. This means that for every 3 moles of Sr(OH)2, we need 2 moles of H3PO4.
Now let's calculate the number of moles of H3PO4 using its concentration and volume:
moles of H3PO4 = concentration of H3PO4 (in M) x volume of H3PO4 (in L)
moles of H3PO4 = 0.30 M x 0.030 L = 0.009 moles
According to the stoichiometry of the balanced equation, we need 2 moles of H3PO4 for every 3 moles of Sr(OH)2, so the number of moles of Sr(OH)2 required is:
moles of Sr(OH)2 = (3 moles Sr(OH)2 / 2 moles H3PO4) x moles of H3PO4
moles of Sr(OH)2 = (3/2) x 0.009 moles = 0.0135 moles
Lastly, we can calculate the volume of 0.10 M Sr(OH)2 using its concentration and the number of moles calculated above:
volume (in L) of Sr(OH)2 = moles of Sr(OH)2 / concentration of Sr(OH)2 (in M)
volume of Sr(OH)2 = 0.0135 moles / 0.10 M = 0.135 L
To convert the volume to milliliters, we multiply the volume in liters by 1000:
volume of Sr(OH)2 = 0.135 L x 1000 mL/L = 135 mL
Therefore, 135 mL of 0.10 M Sr(OH)2 are required to react with 30 mL of 0.30 M H3PO4.
To solve this problem, we need to use the balanced equation provided and apply the concept of stoichiometry.
First, let's identify the stoichiometric ratio between Sr(OH)2 and H3PO4 from the balanced equation:
3Sr(OH)2 + 2H3PO4 → 6H2O + Sr3(PO4)2
From the equation, we see that 3 moles of Sr(OH)2 react with 2 moles of H3PO4.
Next, let's determine the number of moles of H3PO4 in the given solution. To do this, we can use the concentration and volume of the H3PO4 solution:
Moles of H3PO4 = concentration * volume
= 0.30 M * 0.030 L (since 30 mL = 0.030 L)
= 0.009 moles
Since the stoichiometric ratio is 3:2 for Sr(OH)2 and H3PO4, we can calculate the number of moles of Sr(OH)2 required:
Moles of Sr(OH)2 = (3/2) * moles of H3PO4
= (3/2) * 0.009 moles
= 0.0135 moles
Now we can calculate the volume of the 0.10 M Sr(OH)2 solution required to react with 0.0135 moles:
Volume of Sr(OH)2 = moles / concentration
= 0.0135 moles / 0.10 M
= 0.135 L
Since the concentration is given in moles per liter (M), we have the volume of Sr(OH)2 in liters.
To convert the volume to milliliters (mL):
Volume of Sr(OH)2 = 0.135 L * 1000 mL/L
= 135 mL
Therefore, 135 mL of the 0.10 M Sr(OH)2 solution is required to react with 30 mL of the 0.30 M H3PO4 solution.