Question

1) determine the number of milliliters of 4.75 M CBr₂H₂ solution that are needed to make 65.5 ml of 1.05 M of that same solution.

2) determine the number of grams of CaCl₂ that are needed to make a 0.355 liters of a 1.25 Molar solution.

3) 2Al(s) + 6HCl(g) → AlCl₃(aq) + 3H₂(g). Determine the grams of H₂ produced 5.40 grams of Al.
This calculated value is the Theoretical yield. This Theoretical yield is to be used in problem 4.

4) For 5 points, 0.0400 grams of H₂
are actually produced. Determine the Percent yield. Use the answer from problem 12 as the calculated Theoretical Yield.

Answer 1

1.05 M , V=2 1. Si = N₂ S2 . 4.75 M 2 = 65.5 ml, S2 = Nowo, Via V3S2 a TERHADAP Si or, va 65.5 ml x 1.05 M 4.75 M or, v 14.48 ml Cell 2. 1000 ml of a 1.25 (M) Caela = 1.25 moles of = 1.25 * 111 g . = 138.75 g OF, 0.355 L /355m of a 1.25 (4) Cacl2 = 138.75 X 355 1000 = 49.26 g B, 2 Al (s) + 6 Hel (g) → Aldiz (29) + 316 (9). According to the balanced equation, From 2 moles of Al 3 moles of this produced or, 2x26 252 g of AL gives 6 gem of H₂ 1 or 5.40g of Al gives 645.40 52 20.62 g of the 4. I can't understand the asking of the question: