Can anyone solve these Stoichiometry problem with work?

Question

1.How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen?

2. Equation given: 4Na+O2->2Na2O
If you have 100 g of sodium and 60 g of oxygen

A. Find the number of moles of sodium oxide produced.

B. Find the mass of excess reactant left over at the conclusion of the reaction.

Answer 1

1. n(H2) = 50 / (2×1.00) = 25 mol.
n(O2) = 100 / (2×16.00) = 3.125 mol
2H2 + O2 -> 2H2O, so we will use 3.125 mol of O2 and 6.25 mol of H2 to get 6.25 mol of H2O.
m(H2O) = nM = 6.25(2×1.00 + 16.00) = 112.5 g.

2. n(Na) = 100/23.1 = 4.329 mol
n(O2) = 60/32.00 = 1.875 mol
The Na will be completely consumed by 4.329 / 4 = 1.082 mol of oxygen. There is more oxygen that this, so the Na is the limiting reagent. The number of moles of Na2O produced is (2/4) (4.329) = 2.165 mol.
The excess reagent is 1.875 - 1.082 = 0.793 mol of O2, with mass (0.793)(32.0) = 25.4 g.

Answer 2

ans-1. here O2 is the limiting reagent. So, no. of moles of H2O formed = 2(no. of moles of O2) = (100/32)*2
= (200/32)*18 gms = 112.5 gms

ans -2. here Na is the limiting reagent. so no. of moles of na2O formed = 1/2(no. of moles of Na) = 50/23 = around 2.2 moles
excess O2 is left = 60/32 - 25/23

Answer 3

For number 1, you must first create the balanced equation,...
2H2 + O2 -> 2H2O

from this, you see that you need 2 moles of O for every mole of H,...

Calculate the number of moles of each:
mass / grams per mole

O - - 100g/ 16 g/mol = 6.25
H - - 50g/ 1.01g/mol = 49.5

O is the limiting factor, so there will be 6.25 mol of O in the final product, and 12.5 mol of H

100g O + (12.5 mol x 1.01 g/mol)
112.625g

then to 2 sig figs
110 g
or 1.1 x 10^2 g