I need step by step answer for question 1 and 2 Calculate the final temperature
ID: 954935 • Letter: I
Question
I need step by step answer for question 1 and 2 Calculate the final temperature for each of the following mixtures. Assume no heat loss to the environment. Show your method for part b. a. 50.0 g of water at 25.0degreeC is mixed with 50.0 g of water at 85.0degreeC b. 75.0 g of water at 25.0degreeC is mixed with 25.0 g of water at 85.0degreeC. 2. a. In Part B of this experiment, 0.1Og of Mg is added to 50. mL of 1.0M HCI. Which is the limiting reactant? Show calculations. b In Part C, 0.25 g of MgO is added to 50.mL of 1.0M HC1. Which is the limiting reactant?Explanation / Answer
1. a) heat lost by hot water = heat gained by cold water
50*4.18*(x-25) = 50*4.18*(85-x)
x = final temperature = 55 C
b) 75*4.18*(x-25) = 25*4.18*(85-x)
x = final temperature = 40 C
2. Mg(s) + 2HCl(aq) ----> MgCl2(s) + H2(g)
No of mol of Mg = 0.1/24 = 0.00417 mol
No of mol HCl = 50/1000*1 = 0.05 mol
from equation
1 mol Mg = 2 mol HCl
limiting reactant is Mg
b)
MgO(s) + 2HCl(aq) ----> MgCl2(s) + H2O(g)
No of mol of MgO = 0.25/40.3044 = 0.0062 mol
nO OF MOL OF hcL = 50/1000*1 = 0.05 mol
Limiting reactant = MgO