3 g of H2 react with 29 g of O2 to form H2O . Find the amount of the reactant left unreacted?

Question

3 g of H2 react with 29 g of O2 to form H2O . Find the amount of the reactant left unreacted? (A) 6 g (B) 8 g (C) 5.7 g (D) 5.2 g

Tardigrade Admin · Accepted Answer

underset underset= 4.0322 × 2.0162 H2+ underset32 text gO2 arrow underset underset= 36.0322 × (2.016 + 16)2 H2 O 3 g of H2 require O2=(32/4.032)× 3=23.8 text g O2 left unreacted =29-23.8=5.2 text g

Q. 3 g of $H_{2}$ react with 29 g of $O_{2}$ to form $H_{2} O$ . Find the amount of the reactant left unreacted?

6 g

8 g

5.7 g

5.2 g

$= 4.032 2 \times 2.016 2 H_{2} + 32 g O_{2} \to = 36.032 2 \times (2.016 + 16) 2 H_{2} O$
3 g of $H_{2}$ require $O_{2} = \frac{32}{4.032} \times 3 = 23.8 g$
$O_{2}$ left unreacted $= 29 - 23.8 = 5.2 g$

Q. 3 g of H2​ react with 29 g of O2​ to form H2​O . Find the amount of the reactant left unreacted?

6 g

8 g

5.7 g

5.2 g

=4.0322×2.016​2H2​​+32 gO2​​→=36.0322×(2.016+16)​2H2​O​ 3 g of H2​ require O2​=4.03232​×3=23.8 g O2​ left unreacted =29−23.8=5.2 g

Q. 3 g of $H_{2}$ react with 29 g of $O_{2}$ to form $H_{2} O$ . Find the amount of the reactant left unreacted?

$= 4.032 2 \times 2.016 2 H_{2} + 32 g O_{2} \to = 36.032 2 \times (2.016 + 16) 2 H_{2} O$
3 g of $H_{2}$ require $O_{2} = \frac{32}{4.032} \times 3 = 23.8 g$
$O_{2}$ left unreacted $= 29 - 23.8 = 5.2 g$