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The reaction between boron sulfide and carbon is given as:
2B2S3 + 3C → 4B + 3CS2
As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.
Given data:
Mass of C = 2.1 * 10^ 4 g
Mass of B = 3.11*10^4 g
Mass of CS2 = 1.47*10^5
Mass of B2S3 = ?
Now based on the law of conservation of mass:
Mass of B2S3 + mass C = mass of B + mass of CS2
Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5
Mass of B2S3 = 15.7 * 10^4 g
0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 × 119 g = 4.79 g
Answer:
7.335 moles of Cl₂ are required to react with 4.89 miles of Al.
Explanation:
Given data:
Moles of Al = 4.89 mol
Number of moles of Cl₂ required = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Now we will compare the moles of Al and chlorine from balance chemical equation.
Al : Cl₂
2 : 3
4.89 : 3/2×4.89 =7.335 mol
Thus, 7.335 moles of Cl₂ are required to react with 4.89 miles of Al.
