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A metal X forms two different chlorides . If 12.7g of chloride A an 16.3g of chloride B contain 71g and 10.87g of chloride respectively, show that the figures agree with the law of multiple proportion.

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To show that the figures agree with the law of multiple proportion, we need to compare the ratios of the masses of the elements present in the compounds.

Let's assume the metal X has a fixed mass in both chlorides A and B. We can represent this mass as "x."

1. For chloride A:
The mass of chloride A is given as 12.7g, and the mass of chlorine in chloride A is given as 71g. So, we can calculate the mass of metal X in chloride A as follows:
Mass of metal X in chloride A = Mass of chloride A - Mass of chlorine in chloride A
Mass of metal X in chloride A = 12.7g - 71g
Mass of metal X in chloride A = -58.3g

Therefore, the formula of chloride A is XCl(-58.3g). Note that the negative sign indicates that the mass of X is negative, which doesn't make sense physically. We can ignore the negative sign for now.

2. For chloride B:
The mass of chloride B is given as 16.3g, and the mass of chlorine in chloride B is given as 10.87g. So, we can calculate the mass of metal X in chloride B as follows:
Mass of metal X in chloride B = Mass of chloride B - Mass of chlorine in chloride B
Mass of metal X in chloride B = 16.3g - 10.87g
Mass of metal X in chloride B = 5.43g

Therefore, the formula of chloride B is XCl5.43g.

Now, let's compare the ratios of the masses of X and Cl in both chlorides:

Ratio of X to Cl in chloride A = (-58.3g):71g = -0.82:1
Ratio of X to Cl in chloride B = 5.43g:10.87g = 0.5:1

As we can see, the ratios are not in simple whole numbers, which means the figures do not agree with the law of multiple proportion.

Therefore, there might be an error in the given data or calculations. Please recheck the values provided for the masses of chloride A and chloride B, as well as the masses of chlorine in each chloride, to ensure the accuracy of the calculations.