Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N_{2} (g) + H_{2}(g)→ 2NH_{3} (g)

**(i)** Calculate the mass of \(NH_{ 3 }produced\; if \; \times \;10^{ 3 }\) g N_{2} reacts with \(1 \; \times \;10^{ 3 }g \;of\; H_2?\)

**(ii)** Will any of the two reactants remain unreacted?

**(iii)** If yes, which one and what would be its mass.

Asked by Abhisek | 1 year ago | 90

**(i)** Balance the given equation:

\(N_{ 2 }\;(g) \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g) \)

Thus, 1 mole (28 g) of N_{2} reacts with 3 mole (6 g) of H_{2} to give 2 mole (34 g) of \(NH_{ 3 }\)

= \(2 \; \times \;10^{ 3 } g\; of N_2 \;will \;react \;with \)

= \( \dfrac{ 6}{ 28 } \; \times \; 2 \; \times \; 10^{ 3 } g NH_3\)

\(2 \; \times \;10^{ 3 }g\) of N_{2} will react with 428.6 g of H_{2}.

Given:

Amt of H_{2} = \(1 \; \times \;10^{ 3 }\)

28 g of \(N_{ 2 }\) produces 34 g of \(NH_{ 3 }\)

Therefore, mass of\( NH_{ 3 }\) produced by 2000 g of \( NH_{ 2 }\)

= \(\dfrac{ 34 \; g }{ 28 \; g } \; \times \; 2000g\)

**= **2430 g of \(NH_{ 3 }\)

**(ii)** \(H_{ 2 }\) is the excess reagent.

Therefore, \( H_{ 2 }\)will not react.

**(iii)** Mass of H_{2} unreacted

= \(1 \; \times \;10^{ 3 }– 428.6 g\)

= 571.4 g

Answered by Pragya Singh | 1 year agoChlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction:

4 HCl (aq) + MnO_{2}(s) → 2H_{2}O (l) + MnCl_{2}(aq) + Cl_{2} (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction, CaCO_{3} (s) + 2 HCl (aq) → CaCl_{2}(aq) + CO_{2} (g) + H_{2}O(l). What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Find:

**(i)** Empirical formula

**(ii)** Molar mass of the gas, and

**(iii)** Molecular formula

Calculate the number of atoms in each of the following

**(i) **52 moles of Ar

**(ii)** 52 u of He

**(iii)** 52 g of He

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope | Molar mass | Abundance |

\(\, _{ 36 }\textrm{Ar}\) | \( 35.96755 g \; mol^{ -1 }\) | 0.337 % |

\(\, _{ 38 }\textrm{Ar}\) | \( 37.96272 g \; mol^{ -1 }\) | 0.063 % |

\(\, _{ 40 }\textrm{Ar}\) | \( 39.9624g \; mol^{ -1 }\) | 99.600 % |