(a) How many moles of oxygen would be needed to react 58.5 grams of H2 to form water?
Mass of hydrogen = 58.5 grams
Moles of hydrogen 58,5/ 2.016 = 29.02 moles
According to balance equation,
2 moles hydrogen need oxygen = 1 mole
29.02 moles hydrogen need oxygen = 29.02/ 2 = 14.51 moles of oxygen is required for 58.5 grams of hydrogen
(b) How many grams of H2 would be needed to form 120 grams of water?
Mass of water obtained = 120 grams
Moles of water obtained 120/ 18 = 6.67 moles
Now according to balance equation,
2 moles H2O need H2 = 2 moles
6.67 moles H2O need H2 = 2x 6.67/ 2 = 6.67 moles of H2 is required
Grams of hydrogen required = mole x M. weight
= 6.67 X 2.016 = 13 .44 grams of hydrogen is required
Q Given the equation N2 + 3H2 → 2NH3 at STP. How many moles of NH3 would be formed, If 6.3 dm3 of N2 gas react with an excess of hydrogen.?
Volume of I mole of a gas at stp = 22.414 dm3
Volume of Nitrogen given is= 6.3 dm3
Moles of nitrogen = Volume in dm3 = 6.3 dm3 = 0.28125 moles
Molar volume 22.414 dm3
According to balance equation ,
1 mole of N2 give NH3 = 2 moles
0.28125 moles of N2 give NH3 = 2x 0.28125 = 0.5625 moles of NH3 will obtain
1 mole nitrogen give NH3 = 2 mole
Q Calculate the mass of Mg metal required to consume 2560 grams of CO2 in reaction. 2Mg(g) + CO2 (g) → 2MgO(s) + C(s)
Mass of CO2 = 2560 grams
Moles of CO2 = mass of CO2 = 2560 = 58.18 moles
Molar mass 44
According to the balance equation ,
1 mole CO2 need Mg = 2 moles
58.18 moles CO2 need Mg = x = 2x 58.18 = 116.36 moles of Mg is required
Mass of Mg required is = Mole x molar mass = 116.36 x 24 = 2792.73 grams
Q when steam is passed through red hot carbon., a mxture of H2 and CO gases called water gas is formed.
C(s) + H2O(g) → CO(g) + H2(g)
(a) Which is the limiting reagent if 24.5 grams of carbon is mixed with 1.89 moles of water vapors?
Mass of carbon = 24.5 g
Mole of carbon 24.5/12 = 2.04 moles
Mole of water = 1.89 moles
As one mole of water need one mole of carbon, therefore 1.89 moles of water need 1.89 moles of C and we have 2.04 moles of C. therefore water is limiting reactants. While carbon is reactants in excess.
(b) Calculate the amount ( in grams) of the excess reagent left unreacted.
Total amount of carbon = 2.04 mole
Carbon consumed = 1.89 moles
Carbon remained unreacted; 2.04 – 1.89 = 0.15 moles
Mass in grams of carbon = mole x molar mass= 0.15 x 12= 1.8 grams