What mass of water is formed in the reaction of 4.16g H with excess oxygen gas.

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Calculate mole H2 using the formula: n = m/M, where:

n = mole

m = mass (g)

M = molar mass (g/mol)

Calculate molar mass of H2.

M H2 = 2 × 1.008 g/mol = 2.016 g/mol

Calculate moles H2.

n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2

Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.

2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O

The mass of water will be calculated by rearranging the n = m/M formula to isolate m;

m = n × M

Calculate the molar mass H2O.

M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol

Calculate the mass H2O.

m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O

4.16 g H2 with excess O2 will produce 37.2 g H2O.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.