Answer:
Step-by-step explanation:
1. NaOH + H2SO4 → Na2SO4+H2O
The reaction is balanced, so we require 1 mole of each reagent to produce 1 mole of each product.
We have 10.0 g H2SO4 (amu = 98), which will be 5/49 mol of H2SO4
We can see that would also result in 5/49 mol of Na2SO4 as well, according to the reaction.
5/49 mol Na2SO4 (amu = 142) weighs (5/49) * 142 = 14.5 g (rounded to 3 significant figures)
2. 2Al + 3H2O → Al2O3 + 3H2
2 moles of Al reacts with 3 moles of H2O to produce 1 mole of Al2O3 and 3 moles of H2O
We have 5.0 g of Al (amu = 27), which will be 5/27 mol of Al
According to the reaction, that would also result in (5/27*2) mol of Al2O3.
5/54 mol Al2O3 (amu = 102) weighs (5/54) * 102 = 9.44 g (rounded to 3 significant figures)