How many mL of 0.716 M HNO3 are needed to dissolve 7.67 g of MgCO3?
2HNO3(aq) + MgCO3(s) Mg(NO3)2(aq) + H2O(l) + CO2(g)
To find out how many mL of 0.716 M HNO3 are needed to dissolve 7.67 g of MgCO3, we need to use stoichiometry and the concept of molarity.
First, let's calculate the number of moles of MgCO3 using its molar mass.
1. The molar mass of MgCO3 is calculated by adding the atomic masses of each element:
Mg: 24.31 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms)
Total molar mass of MgCO3 = 24.31 + 12.01 + (16.00 x 3) = 84.31 g/mol
2. Now, we can calculate the number of moles of MgCO3:
Moles of MgCO3 = Mass of MgCO3 / Molar mass of MgCO3 = 7.67 g / 84.31 g/mol
Next, we need to determine the stoichiometric ratio between HNO3 and MgCO3 from the balanced equation. The equation tells us that 2 moles of HNO3 react with 1 mole of MgCO3.
3. Since the stoichiometric ratio is 2:1, we can use this ratio to find the number of moles of HNO3 required to react with the given moles of MgCO3:
Moles of HNO3 = (2/1) x Moles of MgCO3
Now, we can use the molarity (M) of HNO3 to calculate the volume (in liters) of HNO3 required to react with the moles of MgCO3 obtained from the previous step.
4. Molarity is defined as moles of solute per liter of solution. In this case, the molarity of HNO3 is 0.716 M, which means there are 0.716 moles of HNO3 in 1 liter of the solution.
Volume of HNO3 (in liters) = Moles of HNO3 / Molarity of HNO3
Finally, we need to convert the volume from liters to milliliters.
5. Volume of HNO3 (in mL) = Volume of HNO3 (in liters) x 1000
By following these steps, you can find how many mL of 0.716 M HNO3 are needed to dissolve 7.67 g of MgCO3.