(3)892)(4) 10 g40 g59. 12 g Mg (at mass 24) react with dilute mineral

1.	(3)892)(4) 10 g40 g59. 12 g Mg (at mass 24) react with dilute mineral acid to produce, (1) 0.5 molmaximum hydrogen equal to(3) 1.5 gVolume at NTP of oxygen required to romnlatobu hurn 1(2) 1.5 mol(4) 0.5g100% carbon)
Answer» lSince the dilute mineral acid is not given, we will use the hydrogen ions: Equation: Mg + 2H^+ -----> Mg (ions) + H2(s) (aq) (s/aq) (g) ratio of moles of Mg: H2 = 1:1 moles of magnesium ions: moles= mass/molar mass 12/24 =0.5 moles if ratio = 1:1 then moles of hydrogen gas produced = 0.5 moles The question requires amount of hydrogen gas:Mass= moles x molar mass =0.5 x 2 =1 gTherefore 1 g of hydrogen gas is produced.

(3)892)(4) 10 g40 g59. 12 g Mg (at mass 24) react with dilute mineral acid to produce, (1) 0.5 molmaximum hydrogen equal to(3) 1.5 gVolume at NTP of oxygen required to romnlatobu hurn 1(2) 1.5 mol(4) 0.5g100% carbon)

Answer»

lSince the dilute mineral acid is not given, we will use the hydrogen ions:

Equation:

Mg + 2H^+ -----> Mg (ions) + H2(s) (aq) (s/aq) (g)

ratio of moles of Mg: H2 = 1:1

moles of magnesium ions: moles= mass/molar mass 12/24 =0.5 moles

if ratio = 1:1 then moles of hydrogen gas produced = 0.5 moles

The question requires amount of hydrogen gas:Mass= moles x molar mass =0.5 x 2 =1 gTherefore 1 g of hydrogen gas is produced.