what will be the law of conversion of a) 65 grams 70 grams

b) magnesium + oxygen = magnesium oxide
45 grams ? grams 80 grams

c) sodium hydroxide + hydrochoride acid
40 grams 36.5 grams
sodium chloride + water
? grams 18 grams

I need help

dzfjhet

To determine the law of conversion for the given reactions, we need to use the concept of stoichiometry. Stoichiometry allows us to calculate the relationship between the amounts of reactants and products in a chemical reaction.

a) To find the law of conversion for the reaction between 65 grams and 70 grams, we first need to determine the balanced chemical equation. Let's assume the reaction is represented as A + B -> C.

To find the molar masses of the reactants, we can consult the periodic table:

- Molar mass of A = 65 g/mol
- Molar mass of B = 70 g/mol

Since we don't have the molar mass of the product, C, we cannot determine the exact amount of C produced. The law of conversion allows us to calculate the relationship between the reactants and products when we have their molar masses.

b) The given reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO) is as follows:

2Mg + O2 -> 2MgO

To determine the law of conversion when 45 grams of Mg reacts, we can use the molar masses:

- Molar mass of Mg = 24.31 g/mol
- Molar mass of O2 = 32.00 g/mol
- Molar mass of MgO = 40.31 g/mol

Using the molar masses and the balanced chemical equation, we can set up the following ratios:

(45 g Mg) / (24.31 g/mol Mg) = (x g MgO) / (40.31 g/mol MgO)
Solving for x, we find:
x = (45 g Mg) * (40.31 g/mol MgO) / (24.31 g/mol Mg)
x ≈ 74.54 g MgO

Therefore, when 45 grams of magnesium reacts, approximately 74.54 grams of magnesium oxide is produced.

To determine the law of conversion when 80 grams of MgO are produced, we can rearrange the equation as follows:

(45 g Mg) / (24.31 g/mol Mg) = (80 g MgO) / (40.31 g/mol MgO)
Solving for x, we find:
x = (80 g MgO) * (24.31 g/mol Mg) / (40.31 g/mol MgO)
x ≈ 48.21 g Mg

Therefore, when 80 grams of magnesium oxide is produced, approximately 48.21 grams of magnesium is consumed.

c) The given reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to form sodium chloride (NaCl) and water (H2O) is as follows:

NaOH + HCl -> NaCl + H2O

To determine the law of conversion for this reaction, we need to know the molar masses:

- Molar mass of NaOH = 40.00 g/mol
- Molar mass of HCl = 36.46 g/mol
- Molar mass of NaCl = 58.44 g/mol
- Molar mass of H2O = 18.02 g/mol

Based on the given amounts, we can set up the following ratios:

(40 g NaOH) / (40.00 g/mol NaOH) = (x g NaCl) / (58.44 g/mol NaCl)
Solving for x, we find:
x = (40 g NaOH) * (58.44 g/mol NaCl) / (40.00 g/mol NaOH)
x ≈ 58.44 g NaCl

Therefore, when 40 grams of sodium hydroxide reacts, approximately 58.44 grams of sodium chloride is produced.

For the production of water:

(40 g NaOH) / (40.00 g/mol NaOH) = (18 g H2O) / (18.02 g/mol H2O)
Solving for x, we find:
x = (40 g NaOH) * (18.02 g/mol H2O) / (40.00 g/mol NaOH)
x ≈ 18.02 g H2O

Therefore, when 40 grams of sodium hydroxide reacts, approximately 18.02 grams of water is produced.