Asample of ammonia reacts with oxygen as shown. 4nh3(g) + 5o2(g) 4no(g) + 6h2o(g) what is the limiting reactant if 4.0 g of nh3 react with 8.0 g of oxygen? o2 because it produces only 0.20 mol of no. nh3 because it produces only 0.20 mol of no. o2 because it produces two times less no than nh3. nh3 because it produces three times more no than o2.

Ans: O2 because it produces only 0.20 mol of NO

Given:

Mass of NH3 = 4.0 g

Mass of O2 = 8.0 g

To determine:

The limiting reagent in the given reaction

4NH3 + 5O2 → 4NO + 6H2O

Explanation:

# moles of NH3 = 4.0/17 g.mol-1 = 0.2353 moles

# moles of O2 = 8.0/32 = 0.25 moles

Based on the reaction stoichiometry:

1 mole of NH3 produces 1 mole of NO

Therefore, 0.2353 moles of NH3 will produce: 0.2353 moles of NO

Similarly, 1 mole of O2 will produce 4/5 moles of NO

0.25 moles of O2 will form: 0.25*4/5 = 0.2 moles of NO

Thus, O2 is the limiting reactant.

O2 because it produces only 0.20 mol of NO.

Explanation:

answer: they produce more slow than unicellular organisms.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.