A sample of InCl3, is known to be contaminated with NaCl. The contaminated sample weighs 2.84g. When the contamintaed sample is treated with excess AgNO3(aq). 574 grams of AgCl(s) are obtained. What is the percentage by mass of InCl3 in the sample? (All I really need to know is the chemical equation that results from this)
Figure the grams of Cl in 574g of AgCl.
Then, that same mass of Cl is in the 2.84g sample.
Let P be the decimalpercent of InCl3 in the sample
Let P1 be the decimal percent of Cl in InCl3 (from the formula)
Let P2 be the decimal percent of Cl in NaCl.
massCl= P*P1*2.84 + (1-P)*P2*2.84
solve for P.
To determine the chemical equation that results from the reaction between the contaminated sample (containing InCl3 and NaCl) and excess AgNO3(aq), we need to consider the relevant reactions and the solubility rules.
AgNO3(aq) and NaCl(aq) can react to form AgCl(s) and NaNO3(aq), according to the following equation:
AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
In this case, the excess AgNO3(aq) reacts with the Cl- ions present in both InCl3 and NaCl to form AgCl(s). The InCl3 does not directly react with AgNO3(aq), but it can still affect the amount of AgCl(s) produced.
Now, let's move on to determining the percentage by mass of InCl3 in the contaminated sample using the given information.
First, we need to find the grams of Cl in 574g of AgCl. Since AgCl is a 1:1 molar ratio between Ag and Cl, we can assume that the mass of Cl in AgCl is the same as the mass of AgCl itself. Therefore, the grams of Cl in 574g of AgCl is also 574g.
Next, we know that the same mass of Cl is in the 2.84g sample. This means that the 2.84g sample contains 574g of Cl.
Now, we can set up an equation to solve for the percentage by mass of InCl3 in the sample. Let P be the decimal percentage of InCl3 in the sample, P1 be the decimal percentage of Cl in InCl3, and P2 be the decimal percentage of Cl in NaCl.
The equation can be written as:
massCl = P * P1 * 2.84 + (1 - P) * P2 * 2.84
We know that the mass of Cl in the sample is 574g, so we substitute this value into the equation:
574g = P * P1 * 2.84 + (1 - P) * P2 * 2.84
Now, you can solve this equation for P to find the percentage by mass of InCl3 in the sample.