**(a) ** **How many moles of oxygen would be needed to react 58.5 grams of H _{2} to form water?**

**Solution;**

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 H _{2} would be needed to form 120 grams of water?**

**Solution;**

Mass of water obtained = 120 grams

Moles of water obtained 120/ 18 = 6.67 moles

Now according to balance equation,

2 moles H_{2}O need H_{2 }= 2 moles

6.67 moles H_{2}O need H_{2} = 2x 6.67/ 2 = 6.67 moles of H_{2} 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 N _{2} + 3H_{2} → 2NH_{3} at STP. How many moles of NH_{3} would be formed, If 6.3 dm^{3} of N_{2} gas react with an excess of hydrogen.?**

** Solution;**

Volume of I mole of a gas at stp = 22.414 dm^{3}

Volume of Nitrogen given is= 6.3 dm^{3}

Moles of nitrogen = __Volume in dm ^{3}__ =

__6.3 dm__= 0.28125 moles

^{3} Molar volume 22.414 dm^{3}

According to balance equation ,

1 mole of N_{2} give NH_{3} = 2 moles

0.28125 moles of N_{2} give NH_{3} = 2x 0.28125 = 0.5625 moles of NH_{3} will obtain

1 mole nitrogen give NH_{3} = 2 mole

**Q Calculate the mass of Mg metal required to consume 2560 grams of CO _{2} in reaction. 2Mg_{(g)} + CO_{2} _{(g) }→ 2MgO_{(s)} + C_{(s)}**

** Solution**;

Mass of CO_{2} = 2560 grams

Moles of CO_{2} = __mass of CO _{2} __=

__2560__= 58.18 moles

Molar mass 44

According to the balance equation ,

1 mole CO_{2 } need Mg = 2 moles

58.18 moles CO_{2} 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 H _{2} and CO gases called water gas is formed.**

** C _{(s)} + H_{2}O_{(g)} → CO_{(g)} + H_{2(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