When 80.0 mL of HCl is added to 80.0 mL of NH3OH, the temperature increases 2.5 celsius. Assuming the final solution has the same specific heat capacity as liquid water, the heat produced in the reaction between two solutions is:
I don't know how to find the moles and mass
80 mL + 80 mL = 160 mL aqueous solution. If you assume the density is the same as that of water;i.e., 1.00 g/mL, than the mass is 160 grams.. The specific heat h2O is 4.184 so
q = 160 x 4.184 x 2.5 = ? in J.
BTW, I think you meant NH4OH (and technically that is NH3(aq)
Oh yes sorry I did mean NH4OH, how do I find the moles?
To find the moles and mass of the substances involved in the reaction, we need to use the equation:
moles = volume (in liters) x molarity
First, let's calculate the moles of HCl and NH3OH used in the reaction:
Given:
Volume of HCl = 80.0 mL = 0.080 L
Volume of NH3OH = 80.0 mL = 0.080 L
Next, we need to know the molarity (M) of each solution. If this information is not provided, we cannot determine the moles and must assume a molarity value:
Let's assume:
Molarity of HCl = 1.0 M (this means there is 1 mole of HCl in 1 liter of solution)
Molarity of NH3OH = 1.0 M (same reasoning as above)
Now we can calculate the moles of each substance:
Moles of HCl = 0.080 L x 1.0 M = 0.080 moles
Moles of NH3OH = 0.080 L x 1.0 M = 0.080 moles
To find the mass of each substance, we need to know the molar mass of each compound. The molar mass values are:
Molar mass of HCl = 36.46 g/mol
Molar mass of NH3OH = 35.04 g/mol
Now we can calculate the mass of each substance:
Mass of HCl = Moles of HCl x Molar mass of HCl = 0.080 moles x 36.46 g/mol = 2.92 grams
Mass of NH3OH = Moles of NH3OH x Molar mass of NH3OH = 0.080 moles x 35.04 g/mol = 2.80 grams
We now have the moles and masses of the substances involved in the reaction.