Question

Consider this reaction 2Mg(s)+O2(g) ———> 2MgO(s) What volume (in milliners) of gas is required to react with 4.03 g Mg at STP ?

A)1860 mL
B)2880 mL
C)3710 mL
D)45,100 mL

Answer 1

The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml

calculation/

• calculate the moles of Mg used

moles=mass/molar mass

moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles

• by use of mole ratio of Mg:O2 from the equation which is 2:1

the moles 02=0.1679 x1/20.0829 moles

• at STP 1 mole of a gas= 22.4 l

0.0895 moles=? L

• by cross multiplication

• =0.0895 moles x22.4 l/ 1 mole=1.8570 L

into Ml = 1.8570 x1000=1856 ml approximately to 1860