Question

1. A student performs the reaction below, which we’ve discussed many times throughout the course. Assume that the student produces 6.101 g of product after performing the reaction on a day where the air pressure was 100.2 kPa and the thermostat was set to 20.0°C. What volume of oxygen gas, in mL, did the reaction use from the classroom’s air? {/12}

2 Mg(s)  +  O2(g) → 2 MgO(s)

 

Answer 1

Product MgO = 6.101 g

6.101 g ×$\frac{1moleMgO}{30.44gMgO}$ = 0.2004 moles MgO

n = 0.2004 moles

P = 100.2kPa

T = 20⁰C = 293K

PV = nRT

100.2×V = 0.2004×0.0821×293

V = 0.04811L

V = 48.11 ml