How many grams of oxygen gas will react completely with a block of calcium metal that is 3.0 cm by 3.5 cm by 4.2 cm, if the density of calcium is 1.55 g/ml?

Answers:

D) 27 g

Step-by-step explanation:

(1) Calculate the mass of Ca

(a) Calculate the volume of the block

V = lwh

V = 4.2 cm × 3.5 cm × 3.0 cm

V = 44.1 cm³

(b) Calculate the mass of the block·

ρ = m/V   Multiply each side by V

m = Vρ

m = 44.1 cm³ × (1.55 g/1 cm³)

m = 68.4 g

(2) Calculate the mass of oxygen

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:  40.08  32.00

          2Ca  +   O₂ ⟶ CaO

m/g:  68.4

(a) Calculate the moles of Ca

n = 68.4 g Ca ×1 mol Ca/40.08 g NH₃

n = 1.71 mol Ca

(b) Calculate the moles of O₂

The molar ratio is (1 mol O₂/2 mol Ca)

n = 1.71 mol Ca × (1 mol O₂/2 mol Ca)

n = 0.853 mol O₂

(c) Calculate the mass of O₂

Mass of O₂ = 0.853 mol O₂ × (32.00 g O₂/1 mol O₂)

Mass of O₂ = 27 g O₂

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

geosphere and hydrosphere.

this is because the geosphere has to do with geological events, while the hydrosphere controls all water on earth.