The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3 g of Mg(HCO3)2 will be

Question

The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3 g of Mg(HCO3)2 will be (A) 2240 mL (B) 1120 mL (C) 2340 mL (D) 2000 mL

Tardigrade Admin · Accepted Answer

Mg(HCO3)2 arrow MgO+2CO2+H2O 146 g of Mg(HCO3)2 gives 22.4 Ã 2 litre of CO2 ⇒ 7.3 g of Mg(HCO3)2 will give (22.4/146)×7.3×2 2.24 L of CO2=2240 mL of CO2 Since, there were 2 molecules of CO thus = 1120 mL

Q. The volume of carbon dioxide gas evolved at S.T.P. by heating $7.3 g$ of $M g (H C O_{3})_{2}$ will be

$2240 m L$

$1120 m L$

$2340 m L$

$2000 m L$

Q. The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3g of Mg(HCO3​)2​ will be

2240mL

1120mL

2340mL

2000mL

Mg(HCO3​)2​→MgO+2CO2​+H2​O 146g of Mg(HCO3​)2​ gives 22.4×2 litre of CO2​ ⇒7.3gofMg(HCO3​)2​ will give 14622.4​×7.3×2 2.24L of CO2​=2240mL of CO2​ Since, there were 2 molecules of CO thus = 1120 mL

Q. The volume of carbon dioxide gas evolved at S.T.P. by heating $7.3 g$ of $M g (H C O_{3})_{2}$ will be

$2240 m L$

$1120 m L$

$2340 m L$

$2000 m L$