How many grams of CaH2CaH2 are needed to generate 147 LL of H2H2 gas if the pressure of H2H2 is 823 torrtorr at 21 ∘C∘C? Express your answer using three significant figures.

139 g of CaH₂ were needed in the reaction

Explanation:

Determine the reaction:

CaH₂ + 2H₂O  →  Ca(OH)₂  +  2H₂

1 mol of calcium hidride reacts with 2 moles of water to produce 1 mol of calcium hydroxide and 2 moles of hydrogen

Let's determine the moles of formed hydrogen by the Ideal Gases Law

P . V = n . R . T

P = 823 Torr . 1 atm/760 Torr = 1.08 atm

T = Absolute T° → T°C + 273 → 21°C + 273 = 294K

1.08 atm . 147 L = n . 0.082 . 294K

(1.08 atm . 147 L) / (0.082 . 294K) = n → 6.60 moles

Ratio is 2:1. We make a rule of three:

2 moles of H₂ came from 1 mol of hydride

Then 6.60 moles of H₂ must came from 3.30 moles of hydride (6.60 .1) /2

Let's convert the moles to mass →  3.30 mol . 42.08 g/ 1 mol =

138.8 g ≅ 139 g

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

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each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i