How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride in the reaction #2AgNO_3 + BaCl_2 -> 2AgCl + Ba(NO_3)_2#?
The equation that is balanced is:
Dimensional analysis can be used to determine the mass of silver chloride when 5.0g of silver nitrate react.
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To find the mass of silver chloride produced, we first need to determine the mole ratio between silver nitrate (AgNO3) and silver chloride (AgCl) from the balanced equation. In this case, the mole ratio is 2:2, or 1:1. Next, we calculate the number of moles of silver nitrate using its molar mass, which is 169.87 g/mol.
5.0 g AgNO3 / 169.87 g/mol = 0.0294 mol AgNO3
Since the mole ratio between AgNO3 and AgCl is 1:1, the number of moles of AgCl produced is also 0.0294 mol. Finally, we find the mass of silver chloride using its molar mass, which is 143.32 g/mol.
0.0294 mol AgCl × 143.32 g/mol = 4.21 g AgCl
Therefore, approximately 4.21 grams of silver chloride are produced.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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