<u>Answer:</u> The volume of barium chlorate is 195.65 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
Given mass of barium chlorate = 25.0 g
Molar mass of barium chlorate = 304.23 g/mol
Molarity of solution = 0.420 mol/L
Volume of solution = ?
Putting values in above equation, we get:
Hence, the volume of barium chlorate is 195.65 mL
Answer:
it's lead (ii) nitrate the name
Answer:
i do not know the answer but pls give a heart and five starts so i can ask questions pls
Explanation:
Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
