ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8

“Magnesium metal is burned in the presence of atmospheric oxygen to form magnesium oxide. Student 1 data Mass Mg 0.1712 g Mass after reaction 0.2886 g Using the given data, calculate the stoichiometric ratio of magnesium ions to oxygen ions in this magnesium oxide sample. Express the ratio as a decimal. Include four significant figures.”

BACKGROUND KNOWLEDGE: The reaction performed is the precipitation of calcium hydroxide from solutions of calcium chloride and potassium hydroxide. The percentage yield of calcium hydroxide is 98.97% QUESTION: Comment on your calculated percentage yield and the main sources of error in this experiment.

If You want to prepare 600 grams of 0.5%sulfuric ácido substance do You need