In order to calculate for the amount of CaCl2 and water used we need to multiply the total amount mixture to the percentage of each component. Since we are given a two component mixture CaCl2-water with 40% CaCl2 by mass, therefore we have 60% water.
Amount of CaCl2 = 0.40*(370.9) =148.36 g of CaCl2
Amount of Water = 0.60*(370.9) = 222.54 g of Water
Answer:
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Explanation:
Answer:
18.7887 g of NaCl
Explanation:
<em>The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?</em>
Let us start by writing out the balanced equation of the reaction:
Na + Cl2 ---> NaCl2
1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.
mole = mass/molar mass
Therefore
18 g of Na = 18/23 = 0.7826 mole
23 g of Cl2 = 23/71 = 0.3239 mole
In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence
0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.
mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g
<u>Hence, 18.7887 grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.</u>
Answer:
9
option d 9
pH of a 10^-5 M HCI solution is 9