2 H2(g) + O2(g) → 2 H2O(g)

How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP?

a

56 L

b

77 L

c

35 L

d

67 L

77 L of water can be made.

Explanation:

Molar mass of = 32 g/mol

So, 55 g of = mol of   = 1.72 mol of  

As hydrogen is present in excess amount therefore   is the limiting reagent.

According to balanced equation, 1 mol of   produces 2 mol of .

So, 1.72 mol of produce mol of or 3.44 mol of .

Let's assume gas behaves ideally at STP.

Then,   , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here, = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So,

Option (b) is correct.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

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