Answer:
pass hahaha dkopa alam yarn
To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
2.5 M x V1 = 1.0 M x .250 L
<span>V1 = 0.10 L or 100 mL of the 2.5 M HCl solution is needed
Hope this helps.</span>
Answer:
CaCO3 is the limiting reactant
55 g of CO2 is made
Explanation:
First we must put down the reaction equation;
CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)
Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles
From the reaction equation;
1 mole of CaCO3 yields 1 mole of CO2
Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2
For HCl;
number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles
From the reaction equation;
2 moles of HCl yields 1 mole of CO2
3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2
Hence CaCO3 is the limiting reactant.
Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2
6.7 mass because 1 atom equals 6.3 but if u add 4 it would be 6.7
