Ex 1:For the combustion of hydrogen:2 H2(g) + O2(g)  2 H2 O(g)a) How much water vapour is produced by 5.0 g of hydrogen?b) What mass of hydrogen is needed to react with 235 g ofoxygen gas?c) What mass of oxygen is required to produce 2.5 moles ofwater vapour?

a) First, we need to find the molar mass of hydrogen (H2), which is approximately 2 g/mol. Therefore, 5.0 g of hydrogen is equivalent to 5.0 g / 2 g/mol = 2.5 moles of hydrogen. According to the balanced chemical equation, 2 moles of hydrogen produce 2 moles of water. Therefore, 2.5 moles of hydrogen will produce 2.5 moles of water. The molar mass of water (H2O) is approximately 18 g/mol, so 2.5 moles of water is equivalent to 2.5 moles * 18 g/mol = 45 g of water.

b) According to the balanced chemical equation, 2 moles of hydrogen react with 1 mole of oxygen. The molar mass of oxygen (O2) is approximately 32 g/mol, so 235 g of oxygen is equivalent to 235 g / 32 g/mol = 7.34 moles of oxygen. Therefore, we need 2 * 7.34 moles = 14.7 moles of hydrogen to react with 235 g of oxygen. The molar mass of hydrogen (H2) is approximately 2 g/mol, so 14.7 moles of hydrogen is equivalent to 14.7 moles * 2 g/mol = 29.4 g of hydrogen.

c) According to the balanced chemical equation, 2 moles of water are produced by 1 mole of oxygen. Therefore, to produce 2.5 moles of water, we need 2.5 moles / 2 = 1.25 moles of oxygen. The molar mass of oxygen (O2) is approximately 32 g/mol, so 1.25 moles of oxygen is equivalent to 1.25 moles * 32 g/mol = 40 g of oxygen.