Question

How many grams of calcium are consumed when 160.6 mL of oxygen gas, measured at STP, reacts with calcium according to the following reaction? 2Ca(s)+O2(g)→2CaO(s)

Answer 1

Calculation of number of moles of oxygen :-

We know that PV = nRT

Where

P = pressure = 1 atm

V = volume = 160.6 mL = 0.1606 L

n = number of moles = ?

R = gas constant = 0.0821 Latm/(mol-K)

T = temperature = 273 K

Plug the values we get

n = (PV) / (RT)

   = (1x0.1606)/ (0.0821x273)

   = 7.165x10^-3 mol

2Ca(s)+O₂(g) → 2CaO(s)

According to the balanced equation,

1 mole of O₂ reacts with 2 moles of Ca

7.165x10^-3 mol of O₂ reacts with 2x7.165x10^-3 mol of Ca

                                                 = 0.0143 mol

Molar mass of Ca = 40 g/mol

We know that number of moles , n = mass/molar mass

So mass of Ca required , m = Number of moles x molar mass

                                         = 0.0143 mol x 40 (g/mol)

                                         = 0.573 g

Therefore the required mass of Ca is 0.573 g