Question

Помогите :) 30 баллов ♡ Какое количество вещества NaOH потребуется для реакции с 5 гр. H2SO4.

Answer 1

2NaOH+H2SO4=Na2SO4+2H2O;
M(H2SO4)=98 г/моль;
n(H2SO4)=m/M=5/98=0,05 моль;
n(NaOH)=2*n(H2SO4)=2*0,05=0,1 моль.

Answer 2

Calculation of the Amount of NaOH Required to React with 5g of H2SO4

To calculate the amount of NaOH required to react with 5g of H2SO4, we need to use the balanced chemical equation for the reaction between NaOH and H2SO4. The balanced equation is as follows:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4. To determine the amount of NaOH required, we need to convert the mass of H2SO4 to moles and then use the mole ratio from the balanced equation.

The molar mass of H2SO4 is calculated as follows: - Hydrogen (H): 2 x 1.008 g/mol = 2.016 g/mol - Sulfur (S): 32.06 g/mol - Oxygen (O): 4 x 16.00 g/mol = 64.00 g/mol

Total molar mass of H2SO4 = 2.016 + 32.06 + 64.00 = 98.076 g/mol

To convert the mass of H2SO4 to moles, we divide the given mass by the molar mass: moles of H2SO4 = 5 g / 98.076 g/mol = 0.051 mol

According to the balanced equation, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, the amount of NaOH required can be calculated as follows: moles of NaOH = 2 x moles of H2SO4 = 2 x 0.051 mol = 0.102 mol

Now, we need to convert the moles of NaOH to grams. The molar mass of NaOH is calculated as follows: - Sodium (Na): 22.99 g/mol - Oxygen (O): 16.00 g/mol - Hydrogen (H): 1.008 g/mol

Total molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol (approximately 40 g/mol)

To convert the moles of NaOH to grams, we multiply the moles by the molar mass: grams of NaOH = moles of NaOH x molar mass of NaOH = 0.102 mol x 40 g/mol = 4.08 g

Therefore, approximately 4.08 grams of NaOH will be required to react with 5 grams of H2SO4.

Please note that the above calculation assumes the reaction goes to completion and that the reactants are in stoichiometric proportions. In practice, the reaction may not go to completion due to various factors, such as impurities or side reactions.